Since this is a right triangle, we can just use Pythagorean Theorem to solve for the side SU:

Length of Hypotenuse^2 = Length of First Leg^2 + Length of Second Leg^2
SU^2 = ST^2 + TU^2
SU = sqrt(ST^2 + TU^2)
SU = sqrt(42^2 + 56^2)
SU = sqrt(4900)
SU = 70

Therefore, the answer is is 70 centimeters.

6 0

3 years ago

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Orly uses 2 cups of raisins for every 8 cups of trail mix she makes. How many cups of trail mix will she make is she uses 12 cup

allsm [11]

You have a 2:8 ratio, or 1:4. Since you have 12 cups of raisins, you're going to get the answer 48.

8 0

3 years ago

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Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago

PLEASE HELP!!!!! I NEED THIS FAST!!!!!!

ioda

Who says V1=V2?
if we simplify we get
(2/3)pir₁³=12pir₂²
for V1 to equal V2

a.
solve for r₁ to find r₁ as a function of r₂
(2/3)pir₁³=12pir₂²
times 3/2 both sides and divide by pi
r₁³=18r₂²
cube root both sides
r₁=∛(18r₂²)

if solve for r₂
(2/3)pir₁³=12pir₂²
divide by 12pi both sides
(1/18)r₁³=r₂²
squer root both sides
√((1/18)r₁³)=r₂

double radius of pond which is r1
√((1/18)r₁³)=r₂
r₁ turns to 2r₁ to double radius
√((1/18)(2r₁)³)=r₂double
√(8(1/18)(r₁)³)=r₂double
(√8)(√((1/18)(r₁)³))=r₂double
√((1/18)r₁³)=r₂ so
(√8)(r₂)=r₂double
(2√2)(r₂)=r₂double
the radius of the tank is multipled by 2√2

5 0

3 years ago