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Katena32 [7]
2 years ago
12

Which relationships describe angles 1 and 2?

Mathematics
1 answer:
Aneli [31]2 years ago
3 0
They are complementary and adjacent angles (Brainliest please)
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NEED ANSWER QUICK WITH STEP BY STEP!!!
Elden [556K]
Y-y(1) = m(x - x(1))

y - 1 = (-3/4)(x + 8)
3 0
3 years ago
120% of what number is 20.4
NeTakaya
20.4÷120=0.17
0.17×100=17
8 0
3 years ago
Stan, a local delivery driver, is paid $3.50 per mile driven pus a daily amount of $75. On Monday, he is assigned a route that i
Alex Ar [27]

To find out how much Stan is paid a day, simply find out how many miles Stan drives per day, and multiply that with the amount Stan is paid per mile (in this case $3.50 per mile). Then, add that with the daily amount Stan earned (which is $75).

For example, on Monday, Stan drove a total of 30 miles (and he's paid $3.50 per mile).

Simply multiply $3.50 (per mile) with 30 (total number of miles driven), and that should equate to $105 earned for driving a total of 30 miles.

Then, add $105 (earned from driving 30 miles) to $75 (daily pay) and that equals $180.

Stan had been paid a total of $180 on Monday.

4 0
3 years ago
If f(x) = 3^+ 10 and g(x) = 2x - 4, find (f - g)(x).
just olya [345]

Answer:

3^x -2x +14

Step-by-step explanation:

I will assume you mean 3^x in the function f(x)

f(x) = 3^x+ 10

g(x) = 2x - 4

(f - g)(x) =  3^x+ 10  - (2x - 4)

       Distribute the minus sign

              =  3^x+ 10  - 2x + 4

              = 3^x -2x +14

8 0
3 years ago
assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28
melomori [17]

Given:

x and y are both differentiable functions of t.

4x^2+3y^3=28

x=-1\text{ and }\dfrac{dy}{dt}=8

To find:

The value of \dfrac{dx}{dt}.

Solution:

We have,

4x^2+3y^3=28       ...(i)

At x=-1,

4(-1)^2+3y^3=28

4+3y^3=28

3y^3=28-4

3y^3=24

Divide both sides by 3.

y^3=8

Taking cube root on both sides.

y=2

So, y=2 at x=-1.

Differentiate (i) with respect to t.

4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0

Putting x=-1, y=2 and \dfrac{dy}{dt}=8, we get

4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0

-8\dfrac{dx}{dt}+9(4)(8)=0

-8(\dfrac{dx}{dt}-9(4))=0

Divide both sides by -8.

\dfrac{dx}{dt}-36=0

\dfrac{dx}{dt}=36

Therefore, the value of 4x^2+3y^3=28 is 36.

6 0
3 years ago
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