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3 years ago

Write an equation for the function that includes the following points (2, 32) and (3, 64)

Mathematics

1 answer:

luda_lava [24]3 years ago

6 0

Answer:

y=32x-32

Step-by-step explanation:

this time no algorithm or equation was needed i saw that if the equation was y=32x there would be ordered pairs (1,32) and (2,64) so to delay it by 1 on the x side just subtract the slope from the y - intercept sorry if that doesn't make sense

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What is the solution of each equation? <br><br> 2s = 34

scZoUnD [109]

Answer:

Step-by-step explanation:

sdfbh

3 0

3 years ago

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is

sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0

3 years ago

????????????????????????????????????????????????????????

defon

Answer:

The last one

Step-by-step explanation:

0.004*.004=0.0016 and k*-K=K squared

4 0

3 years ago

Read 2 more answers

A convex heptagon has interior angels of 125, 155, 95, 135, 120, and 115. What is the measure of the 7th angle?

wel

Sum of those 6 angles is 745 degrees
7th angle should make it equal 900
900-745=155

3 0

4 years ago

Help me please !! i need the answer

madam [21]

Answer:

$PQ=22\\\\QR=7\\\\PR=29$

Step-by-step explanation:

To find the value of y, you can create an equation where two parts of the line are equal to the whole line:

$PQ+QR=PR$

Substitute given values:

$(8y+6)+(y+5)=(19y-9)$

Solve for y. Simplify parentheses:

$8y+6+y+5=19y-9$

Combine like terms to simplify the equation:

$8y+y+6+5=19y-9\\\\9y+11=19y-9$

Get the variable on one side of the equation and the constants on the other. Add 9 to both sides:

$9y+11+9=19y-9+9\\\\9y+20=19y$

Subtract 9y from both sides:

$9y-9y+20=19y-9y\\\\20=10y$

Isolate the variable. Divide both sides by 10:

$\frac{20}{10} =\frac{10y}{10} \\\\y=2$

The value of y is 2. Insert this value into each line segment value and solve:

$PQ=8(2)+6\\\\PQ=16+6\\\\PQ=22$

$QR=(2)+5\\\\QR=7$

$PR=19(2)-9\\\\PR=38-9\\\\PR=29$

:Done

You can check by using:

$PQ+QR=PR\\\\22+7=29$

This statement is true, so the values are correct.

7 0

3 years ago