Which expression is equivalent to 7^3 ⋅ 7^−5? A.7^2 B.7^7 C.1 over 7 to the 2nd power D.1 over 7 to the 7th power

The coordinates of trapezoid EFGH are E(4, 8), F(1, 2), G(6, 1), and H(8, 8). The image of EFGH under dilation is E'F'G'H'. If t

SSSSS [86.1K]

The image of EFGH under dilation is E'F'G'H'. If the coordnates of Vertex H' are (4,4), the coordinates of vertex E' are (2,4)

Second option (2,4)

Solution:

The point H has coordinates (8,8)

H=(8,8)=(xh,yh)→xh=8, yh=8

If the coordinates of vertex H' are (4,4)

H'=(4,4)=(xh',yh')→xh'=4, yh'=4

The factor of dilation "f" is:

f=xh'/xh=yh'/yh

f=4/8=4/8

f=4/8

Simplifying the fraction, dividing the numerator and denominator by 4:

f=(4/4)/(8/4)

f=1/2

Then we must multiply the original coordinates of EFGH by f=1/2 to obtain the new coordinates E'F'G'H'. Then, the coordinates of E' are:

E=(4,8)=(xe,ye)→xe=4, ye=8

xe'=f xe→xe'=(1/2) 4→xe'=4/2→xe'=2

ye'=f ye→ye'=(1/2) 8→ye'=8/2→ye'=4

E'=(xe',ye')→E'=(2,4)

6 0

4 years ago

The velocity function, in feet per second, is given for a particle moving along a straight line. Find (a) the displacement and (

AveGali [126]

Answer:

(a) 2 feet.

(b) 2 feet.

Step-by-step explanation:

We have been given that the velocity function $v(t)=\frac{1}{\sqrt{t}}$ in feet per second, is given for a particle moving along a straight line.

(a) We are asked to find the displacement over the interval $1\leq t\leq 4$ .

Since velocity is derivative of position function , so to find the displacement (position shift) from the velocity function, we need to integrate the velocity function.

$\int\limits^b_a {v(t)} \, dt$

$\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$

$\int\limits^4_1 {\frac{1}{t^{\frac{1}{2}}} \, dt$

$\int\limits^4_1 t^{-\frac{1}{2}} \, dt$

Using power rule, we will get:

$\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}}\right] ^4_1$

$\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}\right] ^4_1$

$\left[2t^{\frac{1}{2}}\right] ^4_1$

$2(4)^{\frac{1}{2}}-2(1)^{\frac{1}{2}}=2(2)-2=4-2=2$

Therefore, the total displacement on the interval $1\leq t\leq 4$ would be 2 feet.

(b). For distance we need to integrate the absolute value of the velocity function.

$\int\limits^b_a |{v(t)|} \, dt$

$\int\limits^4_1 |{\frac{1}{\sqrt{t}}}| \, dt$

Since square root is not defined for negative numbers, so our integral would be $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ .

We already figured out that the value of $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ is 2 feet, therefore, the total distance over the interval $1\leq t\leq 4$ would be 2 feet.

7 0

3 years ago

Suppose the American Lung Association stated that approximately 1 out of every 20 deaths in the US last year was from lung cance

tester [92]

Answer: 10

Step-by-step explanation:

Given : The American Lung Association stated that approximately 1 out of every 20 deaths in the US last year was from lung cancer.

Then, the proportion of death will be :_

$p=\dfrac{1}{20}=0.05$

Now, If there are 200 students in a class, then the expected number of persons die from lung cancer :-

$p\times200\\\\=(0.050)\times200=10$

Hence, we expect the number of persons die from lung cancer =10

8 0

4 years ago

In a School there are 60 teachers and 1200 students.

alexdok [17]

369

60(0.15)=9
1200(0.3)=360

6 0

4 years ago

Read 2 more answers

What fraction of the months of the year have 31 days

Dmitry_Shevchenko [17]

Well 7 out of 12 months have 31 days and if you need to know what the names are they are ........January,March,May,July,August,October, and December hope this helps!

3 0

4 years ago