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3 years ago

I need help or no points :)

Mathematics

2 answers:

frez [133]3 years ago

7 0

Answer:

1 and 9

Step-by-step explanation:

Lady bird [3.3K]3 years ago

3 0

X intercept is (17,0)
Y intercept is ( , ) look in picture
I don’t know if it’s right but hope it is

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Can anybody figure this question out cause I can't figure it out.

Kazeer [188]

Answer:

there is no change because it is a circle

4 0

3 years ago

Solve the system by the elimination method <br> 7x+5y=-25<br> 5x+3y=-23

aksik [14]

Step-by-step explanation:

7x+5y=25 ×3 7×-1.5+5y=25

5x+3y= -23 ×5 -10.5 +5y =25

5y= 25 + 10.5

21x+15y=75 5y= 35.5

25x+15y=69 y= 7.1

-4x=6

x=-1.5

5 0

4 years ago

PLS HELP, i am literally crying over this, Guaranteed brainliest if you get it right

Agata [3.3K]

I’m pretty sure the answer is -10

4 0

3 years ago

Read 2 more answers

Guys look at the pic of the park and read the question I need this ASAP WITH ALL STEPS the choices are

Alik [6]

Answer:

A. 821.5 Square Yards

Step-by-step explanation:

30 x 30= 900

900 is the area for the whole park

To find the pool area you need to do pi times Radius squared

so 3.14 x 5 x 5

That equals 78.5 for the pool area.

So to find the grass area you need to subtract the pool area from the park

900 - 78.5 = 821.5 square yards!

5 0

3 years ago

Solve the initial value problem

tigry1 [53]

$9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}$

Look for an integrating factor $\mu(t)$ :

$\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}$

Multiply both sides by $\mu$ :

$(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}$

Condense the left side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}$

Integrate both sides:

$(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt$

For the integral on the right, substitute

$u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du$

$\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du$

$\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C$

$\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)$

$\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C$

$\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}$

Given that $y(0)=12$ , we get

$12=9+C\implies C=3$

$\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}$

6 0

3 years ago

I need help or no points :)​

I need help or no points :)