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3 years ago

Can sumone help please!

Mathematics

1 answer:

balu736 [363]3 years ago

7 0

Thr answer is B) y= -4x + 48
since if you replace x with 2, you would get 40 when you solve the answer.

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What is the answer to: x + y = 4 Y= x+4 Using elimination

Ket [755]

X + y = 4
y = x + 4

x + (x + 4) = 4
y = x + 4

2x + 4 = 4
y = x + 4

2x = 4 - 4
y = x + 4

x = 0
y = 0 + 4

x = 0
y = 4

5 0

3 years ago

4х – бу =11 3х + 2y = 9

k0ka [10]

Answer:

(3/26, 38/13)

Step-by-step explanation:

I hope I solved it right and it helped you somehow

If you have any questions about the way I solved it, don't hesitate to ask :)

5 0

3 years ago

The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.

klemol [59]

Answer:

A: $v(t)=-t^2+4t-5$

Step-by-step explanation:

Acceleration is second derivative of position, velocity is first derivative. Therefore, the velocity is the integral of acceleration.

$v(t)=\int\ {2t+4} \, dt$

Integrate:

$-t^2+4t+C$

V(0)=-5:

$-0^2+4(0)+C=-5\\C=-5$

Therefore, v(t):

$v(t)=-t^2+4t-5$

7 0

3 years ago

Who knows how to do this ughhh I need help!

suter [353]

Answer:

$\large\boxed{\sqrt{56x^{17}}=2x^8\sqrt{14x}}$

Step-by-step explanation:

$Domain:\ x\geq0\\\\\sqrt{56x^{17}}=\sqrt{4\cdot14\cdot x^{16+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{4\cdot14\cdot x^{16}\cdot x^1}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt4\cdot\sqrt{14}\cdot\sqrt{x^{16}}\cdot\sqrt{x}=2\cdot\sqrt{14}\cdot\sqrt{x^{8\cdot2}}\cdot\sqrt{x}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=2\cdot\sqrt{14}\cdot\sqrt{(x^8)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2\cdot\sqrt{14}\cdot x^8\cdot\sqrt{x}=2x^8\sqrt{14x}$

8 0

3 years ago

The ratio of dogs to cats is 2 to 3. There are 15 pets total. How many dogs are there? How many cats are there?

ElenaW [278]

So there are 2 dogs to every 3 cats (put into words)

in order to reach 15 total multiply both 2 and 3 by 3

2x3 = 6
3x3 = 9

so there are 6 dogs and 9 cats

6 0

3 years ago