3 years ago

Can someone help with this question please??

Mathematics

1 answer:

jonny [76]3 years ago

7 0

A) P = -3, G = 3, -G = -3, P = -G.

You might be interested in

For the problem I tried dividing but my answers were not correct. How can I solve this problem then? Can someone help me out her

astraxan [27]

Answer:

Step-by-step explanation:

5 = 40

1 = x

Then we multiply by the rule of crisscrossing

5 x X = 40 x 1

5x = 40 then divide both by 5

X = 8

3 0

3 years ago

Hunter is 1.75 meters tall. At 12 noon, he measures the length of a tree's shadow to be 30.85 meters. He stands 26.8 meters away

Likurg_2 [28]

Enjoy mu little post-it note drawings

7 0

3 years ago

Read 2 more answers

Julia drinks 16 cups of water per day. How many quart(s) of water does she drink per day?

ivann1987 [24]

4 cups are in a quart. Therefore you must use divide the amount of cups by 4 to get an amount of quarts. therefore you have 4 quarts.

4 0

4 years ago

Read 2 more answers

Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx

shutvik [7]

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$ , so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$ :

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

$\csc v-\cot v=x+C$

$\csc(x+y)-\cot(x+y)=x+C$

Given that $y(0)=\dfrac\pi2$ , we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

$\csc(x+y)-\cot(x+y)=x+1$

5 0

3 years ago

Please help.<br> Algebra.

blsea [12.9K]

Answer:

Question 5: A Question 6: B

6 0

3 years ago