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3 years ago

12

GEOMETRY. I will give brainliest

2 answers:

spayn [35]3 years ago

5 0

Answer:

48 is my best guess

Step-by-step explanation:

Andreas93 [3]3 years ago

4 0

Answer:

Step-by-step explanation: 24

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The product of 5 and m squared, increased by the sum of the square of m and 5

weqwewe [10]

$The\ product\ of\ 5\ and\ m=5\times m=5m\\\\increased\to+\\\\the\ sum\ of\ a\ square\ of\ m\ and\ 5=m^2+5\\\\\huge\boxed{5m+(m^2+5)=5m+m^2+5}$

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4 years ago

Can you pls help me it’s timed

Lerok [7]

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The median is 22

Step-by-step explanation:

6 0

3 years ago

Is there a multiple of 3? And why not or why yes?

rodikova [14]

Yes there is because u can multiply it by different numbers except for 1 and 0

6 0

3 years ago

Read 2 more answers

What is the slope of the line determined by the points (5,-3) and (-9,-6)

Serggg [28]

Answer:

slope = $\frac{3}{14}$

Step-by-step explanation:

Calculate the slope m using the slope formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (5, - 3) and (x₂, y₂ ) = (- 9, - 6)

m = $\frac{-6+3}{-9-5}$ = $\frac{-3}{-14}$ = $\frac{3}{14}$

6 0

3 years ago

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

3 years ago