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3 years ago

14

Sharon went bowling with her friends. She paid $3 to rent

1 answer:

gavmur [86]3 years ago

4 0

Answer:

3 + 3.75x=18

subtract 3 from both sides

3.75x=15

divide by 3.75 on both sides

x=4

Step-by-step explanation:

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2. HELP ASAP!!! GIVING BRAINLIEST!!!<br> Find each missing measure.

katrin2010 [14]

3. 49
4. 26

u just add up all the sides then subtract it from 180

8 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tan

valina [46]

$2(x^2+y^2)^2=25(x^2-y^2)$

Let $y=y(x)$ , so that differentiating both sides wrt $x$ gives

$4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)$

If $x=-3$ and $y=1$ , the above reduces to

$40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}$

This is the slope of the tangent line, which has equation

$y-1=\dfrac9{13}(x+3)\implies\boxed{y=\dfrac{9x+40}{13}}$

7 0

3 years ago

Enter the y coordinate of the solution to this system of equations -2x+3y=-6. 5x-6y=15

adoni [48]

The intersect of the 2 lines is (3,0) and the y coordinate is 0

4 0

4 years ago

A 4-foot tall child walks directly away from a 12-foot tall lamppost at 2 mph. How quickly is the length of her shadow increasin

xz_007 [3.2K]

Answer:

The length of the shadow is increasing with the rate of 1.5 feet per sec

Step-by-step explanation:

Let AB and CD represents the height of the lamppost and child respectively ( shown below )

Also, let E be a point represents the position of child.

In triangles ABE and CDE,

$\angle ABE\cong \angle CDE$ ( right angles )

$\angle AEB\cong \angle CED$ ( common angles )

By AA similarity postulate,

$\triangle ABE\sim \triangle CDE$

∵ Corresponding sides of similar triangles are in same proportion,

$\implies \frac{AB}{CD}=\frac{BE}{DE}$

We have, AB = 12 ft, CD = 4 ft, BE = BD + DE = 6 + DE,

$\implies \frac{12}{4}=\frac{6+DE}{DE}$

$12DE = 24 + 4DE$

$8DE = 24$

$DE=3$

Now, the speed of walking = 2 mph = $\frac{2\times 5280}{3600}\approx 2.933\text{ ft per sec}$

Note: 1 mile = 5280 ft, 1 hour = 3600 sec

Thus, the time taken by child to reach at E

$= \frac{\text{Walked distance}}{\text{Walking speed}}$

$=\frac{6}{2.933}$

= 2.045 hours

Hence, the change rate in the length of shadow

$= \frac{\text{Length of shadow}}{\text{Time taken}}$

$=\frac{3}{2.045}$

= 1.5 ft per sec.

3 0

3 years ago

A hole of radius 6cm is drilled through a right circular cylinder whose base radius and height are 9cm and 12cm respectively. Wh

eduard

Answer:

CSA of cylinder= 2πr(h+r)

=2 π×9 (21)

= 378π cm²

now

CSA of hole= 2πr

12π cm²

7 0

3 years ago