Answer:
There are 4 terms
Step-by-step explanation:
A term is a single mathematical expression. Terms can be identified with a plus or minus sign in front of them. Terms can also be multiplied and divided.
So, in this case, the terms are:
-7
12x^4
-5y^8
x
the first day she walked 27 kilometers. each day since she walked 2/3 of what she walked the day before. what is the total distance cecelia has traveled be the end of the 5th day?
Answer: 70
Step-by-step explanation:
We are required to calculate the total distance Cecilia travelled in 5 days
The total distance Cecilia travelled for 5 days is 99 kilometers
Day 1 = 27 kilometers
Day 2 to day 5 = 2/3 of 27
= 2/3 × 27
= 2 × 9
= 18 kilometers each day
Total distance = day 1 + day 2 + day 3 + day 4 + day 5
= (27 + 18 + 18 + 18 + 18) kilometers
= 99 kilometers
Therefore, the total distance Cecilia travelled for 5 days = 99 kilometers
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The weights of certain machine components are normally distributed with a mean of 8.04 g and a standard deviation of 0.08 g. Find the two weights that separate the top 3% and the bottom 3%. (These weights could serve as limits used to identify which components should be rejected)
Answer:
The bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.
Step-by-step explanation:
We are given that
Mean, [tex]\mu=8.04 g[/tex]
Standard deviation, [tex]\sigma=0.08g[/tex]
We have to find the two weights that separate the top 3% and the bottom 3%.
Let x be the weight of machine components
[tex]P(X<x_1)=0.03, P(X>x_2)=0.03[/tex]
[tex]P(X<x_1)=P(Z<\frac{x_1-8.04}{0.08})[/tex]
=0.03
From z- table we get
[tex]P(Z<-1.88)=0.03, P(Z>1.88)=0.03[/tex]
Therefore, we get
[tex]\frac{x_1-8.04}{0.08}=-1.88[/tex]
[tex]x_1-8.04=-1.88\times 0.08[/tex]
[tex]x_1=-1.88\times 0.08+8.04[/tex]
[tex]x_1=7.8896[/tex]
[tex]\frac{x_2-8.04}{0.08}=1.88[/tex]
[tex]x_2=1.88\times 0.08+8.04[/tex]
[tex]x_2=8.1904[/tex]
Hence, the bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.
Uma empresa do ramo de confecções produz e comercializa calças jeans. Se x representa a quantidade produzida e comercializada (em milhares de reais) e L(x) = -x² + 8x – 16 representa o lucro (em milhares de reais) da empresa para x unidades, então quando L(x) = 0 a empresa terá produzido e comercializadas quantas unidades dessas calças jeans? *
Answer:
I am in 5 classso i did not known answer
One leg of a right triangle is 7 inches longer than the other leg, and the hypotenuse is 35 inches. Find the lengths of the legs of the triangle.
Answer: 21, 28
Step-by-step explanation:
Side #1 = xSide #2 = x + 7Hypotenuse = 35Use the Pythagorean Theorem [tex]a^{2}+b^{2}=c^{2}[/tex]:
a = xb = x + 7c = 35Substitute in the values & solve:
[tex]x^{2}+(x+7)^{2}=35^{2}\\x^{2}+x^{2}+14x+49=1225\\2x^{2}+14x+49-1225=0\\2x^{2}+14x-1176=0\\2(x^{2}+7x-588)=0\\2(x + 28)(x - 21)=0\\x_{1}=-28, x_{2}=21[/tex]
-28 is not a possible solution since you can't have negative inches...
a = x = 21b = x + 7 = 21 + 7 = 28c = 35In one week, Alex worked 35 hours at $ 12.50 per hour plus 5 hours overtime at 'time and half'. How much would Alex have earned for the week? *
Find the value please!!!
the answer is in the picture
PLEASE ANSWER ASAAPPPPPPPP
Answer:
3
Step-by-step explanation:
Question 4 of 10
Which describes the graph of y=-(x - 3)^2 + 2?
O A. Vertex at (-3,2)
O B. Vertex at (3,-2)
C. Vertex at (3,2)
O D. Vertex at (-3,-2)
Answer:Vertex at (3, -2)
Step-by-step explanation:
If f(x) = 6x - 4. what is f(x) when x = 8?
Answer:
44
Step-by-step explanation:
* means multiply
put 8 where the x is
6 * 8 - 4
48 - 4
44
Find x and y
Help me please
Answer:
x = 70 y = 140
Step-by-step explanation:
Apologies if I get any names of vocabulary terms wrong, I never pay much attention to the names.
The central angle measure of an arc is the same as the arc's measure. In this case, y is the central angle and 140 is the arc's measure, so y = 140.
The inscribed angle measure of an arc is 1/2 the arc's measure. An inscribed angle lies on the circle. So 140/2 = 70 = x.
A jet travels 832 km in 5 hours.at this rate,how far could the jet fly in 12 hours? What is the rate of speed of the jet
Answer:
1344.
Step-by-step explanation :
.
The speed of the jet will be 166.4. If the jet fly for 12 hours, then the distance cover will be 1996.8.
What is speed?The distance covered by the particle or the body in an hour is called speed. It is a scalar quantity. It is the ratio of distance to time.
We know that the speed formula
Speed = Distance/Time
A jet travels 832 km in 5 hours.
Then the speed of the jet will be
Speed = 832 / 5
Speed = 166.4 km per hours
If the jet fly for 12 hours, then the distance cover will be
Distance = Speed x Time
Distance = 166.4 x 12
Distance = 1996.8 km
More about the speed link is given below.
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PLEASE HELP!!! Urgent. Will mark as brainliest
Answer:
Step-by-step explanation:
m∠2 = [tex]\frac{1}{2}[/tex](55+41)
m∠2 = [tex]\frac{1}{2}[/tex](96)
m∠2 =48°
m∠1 = 180-48 = 132°
Write the expression as an exponent:
5^8 x 25
6^15 x 36
Answer:
a.) 5¹⁰
b.) 6¹⁷
Step-by-step explanation:
a.) 5⁸ × 25
Write 25 in the exponential form with the base of 5.
5⁸ × 5²To calculate product use exponent rule
5⁸+²5¹⁰b.) 6¹⁵ × 36
Similarly, 6¹⁵ × 36
Write 25 in the exponential form with the base of 6.
6¹⁵ × 6²To calculate product use exponent rule.
6¹⁵ + ²6¹⁷Answer:
[tex] \displaystyle {5}^{10} [/tex]
[tex] \displaystyle {6}^{17} [/tex]
Step-by-step explanation:
Question-1:we want to rewrite the following expression as an exponent
[tex] \displaystyle {5}^{8} \times {25}[/tex]
remember that 25 is the square of 5 therefore
[tex] \displaystyle {5}^{8} \times {5}^{2} [/tex]
recall that,
[tex] \displaystyle {x}^{m} \times {x}^{n} = {x}^{m + n} [/tex]
with that law we obtain:
[tex] \displaystyle {5}^{8 + 2} [/tex]
simplify addition:
[tex] \displaystyle {5}^{10} [/tex]
Question-2:likewise Question-1 36 is the square of 6 Thus,
[tex] \displaystyle {6}^{15} \times {6}^{2} [/tex]
similarly apply law of exponent:
[tex] \displaystyle {6}^{15 + 2} [/tex]
simplify addition:
[tex] \displaystyle {6}^{17} [/tex]
hence,
we have written the expression as an exponent
How long is JK?
Choose 1 answer:
A: 3
B: 2 square root 3
C: 3 square root 3
D: 6
E: 9
Answer:
C) 3 times the sq root of 3
Step-by-step explanation:
In a right angle triangle, the COS is definned as the ajacent side over the hypotenuse. So
COS 60 = JK/JL
COS 60 = JK/(6*sq root 3)
.5 = JK/(6*3**1/2)
JK = .5 *(6*3**1/2)
JK = 3*3**1/2
what is the length of segment SR?
ST = ST (Common)
RT = TQ (Given)
∠ RTS = ∠ SQT (Both 90°)
So, ∆ RTS ≅ ∆ QTS (By SAS congruency rule)
So, RS = QS (By C.P.C.T.)
=> 2x + 8 = 8x - 4
=> 8 + 4 = 8x - 2x
=> 12 = 6x
=> 12/6 = x
=> 2 = x
SR = 2x + 8 (Given)
=> SR = 2(2) + 8 (Putting the value of x)
=> SR = 4 + 8
=> SR = 12
Answer:
Step-by-step explanation:
Can someone please help me really struggling
Answer:
your y-intercept is 5 and your slope is 1 hope this helps with translating into the form you want
Step-by-step explanation:
Please helpppp
Find l.
Answer:
Step-by-step explanation:
radius of base=8/2=4 ft
l²=4²+9²=16+81=97
l=√97 ft
Find the missing side or angle. Round to the nearest tenth. C=95° a=5 c=6 A=[?]°
Answer:
Missing angle A = 56.1° (approximately)
Step-by-step explanation:
<A = arcsin(a×sin(C)/c)
= arcsin(5×sin 95/6)
= 56.1154084° ≈ 56.1°
Answered by GAUTHMATH
if p-1/p = 4 Find (p+1/p)²
Answer:
4
Step-by-step explanation:
[tex] \frac{p - 1}{p} = 4[/tex]
Solve for p.
The numerator. difference has to be 4 times the answer of the denomiator.
So we can say
[tex]p - 1 = 4p[/tex]
Solve for p
[tex] - 1 = 3p[/tex]
[tex]p = - \frac{1}{3} [/tex]
Now plug this in for the other formula.
[tex]( \frac{ - \frac{1}{3} +1}{ \frac{1}{3} } ) {}^{2} = 2 {}^{2} = 4[/tex]
So
Log problem below in the picture
Your answer is 2.98004491789381.
A car and a bus were travelling towards each other at uniform speeds. They were 250 km apart at noon and passed each other at 2 p.m. If the speed of the car was 75 km/h, find the speed of the bus.
Answer:
50 km/hr
Step-by-step explanation:
speed = x + 75
distance = 250km
time = 2
250 = (x + 75) * 2
125 = x + 75
x = 50
out of 1300 students in a school , only 60% passed . find the number of students failed .
Hi there!
»»————- ★ ————-««
I believe your answer is:
520 students.
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
60% of the 1300 students passed. This means that 40% of the students failed.
⸻⸻⸻⸻
[tex]1300 * 0.4= 520[/tex]
⸻⸻⸻⸻
520 students failed school.
⸻⸻⸻⸻
[tex]\boxed{\text{Another way to find this...}}\\\\\text{Take 60}\% \text{ of 1300:}\\\\1300 * 0.6 = 780\\\\\text{780 students passed school.}\\\\\text{Subtract 780 from 1300:}\\\\1300 - 780 = \boxed{520}[/tex]
⸻⸻⸻⸻
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
help.........................
Answer:
solution
here,
Step-by-step explanation:
f={(2,1/2), ( 3, 1/3) , (4, 1/4)}
Range = { / (\ frac 12\) , \ (\ frac 13\) ,
\( \ frac 14\ )}
Inverse function ( f-^1 ) = {( 1/2, 2) ,( 1/3 ,3) , 1/4, 4 )} is the required answer
Find angle D if angle B = 50
============================================================
Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
----------------------------
Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
---------------------------
Here's what we have so far for quadrilateral DAEC
angle A = 90angle E = 100angle C = 90angle D = unknownNow we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.
What is the equation of the following line?
(-4, 8)
(0,0)
Answer
Step-by-step explanation:
y = -2x + 0
Therefore; y = -2x
What percentage of temperatures are below 55°? A. 50% B. 25% C. 75% D. 20%
Step-by-step explanation:
C
im not sure of my answer. hope to help
Are the two triangles similar. If so, State how
Answer:
Yes by AAA
Step-by-step explanation:
Angle A is the same for both triangles. Angle D is congruent to Angle A since BC and DE are parallel lines and same with angle E. So by AAA the triangles are similar
In a geometric progression, the first tern is 81 and the fourth term is 24. find the common ratio
Step-by-step explanation:
81/24= 27/8Hello,
Let's say a the first term and r the ratio of the G.P.
[tex]u_1=a=81\\u_2=a*r\\u_3=a*r^2\\u_4=a*r^3\\\\\dfrac{u_4}{u_1} =\dfrac{a*r^3}{a} =r^3=\dfrac{24}{81} =\dfrac{8}{27} \\\\r=\sqrt[3]{\dfrac{8}{27} } \\\\\boxed{r=\dfrac{2}{3}}\\[/tex]
Need help on this!!!!
Answer:
D
Step-by-step explanation:
First find h(3)
2(3)-2= 4
then find g(4)
4³-5= 59
If A = [tex]\left[\begin{array}{ccc}cosx&-sinx\\sinx&cosx\end{array}\right][/tex], then show that [tex](A^{-1} )^{-1}[/tex]
Given:
The matrix is:
[tex]A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}[/tex]
To show:
[tex](A^{-1})^{-1}=A[/tex]
Solution:
If a matrix is:
[tex]M=\begin{bmatrix}a&b\\c&d\end{bmatrix}[/tex]
Then,
[tex]M^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}[/tex]
We have,
[tex]A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}[/tex]
Using the above formula, we get
[tex]A^{-1}=\dfrac{1}{(\cos x)(\cos x)-(-\sin x)(\sin x)}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}[/tex]
[tex]A^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}[/tex]
[tex]A^{-1}=\dfrac{1}{1}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}[/tex]
[tex]A^{-1}=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}[/tex]
Now, the inverse of [tex]A^{-1}[/tex] is:
[tex](A^{-1})^{-1}=\dfrac{1}{(\cos x)(\cos x)-(\sin x)(-\sin x)}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}[/tex]
[tex](A^{-1})^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}[/tex]
[tex](A^{-1})^{-1}=\dfrac{1}{1}A[/tex]
[tex](A^{-1})^{-1}=A[/tex]
Hence proved.