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puteri [66]
3 years ago
14

Need help figuring this one out

Mathematics
2 answers:
Usimov [2.4K]3 years ago
7 0
What the person above me said lol ^^ i was going to say the exact same thing
Afina-wow [57]3 years ago
6 0

Answer:

A= none

B= none

C= vertical angels, complementary and supplementary angels.

D= Adjacent angels

Step-by-step explanation:

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F(x)=-4x - 10 find x when f(x)=10
BARSIC [14]

Answer:

-40

Step-by-step explanation:

f(10)=-4(10)

f(10)=-40

4 0
3 years ago
NiCole Spent 15% of her day doing homework. How many hours did she spend on her homework?
Whitepunk [10]

Answer:

4 hour of her day

Step-by-step explanation:

4 0
2 years ago
If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,
S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
3 0
3 years ago
You work at a bike shop. A customer purchases 6 new bike tires at $6.99 each. If the customer gives you $40, how much is still o
Mariulka [41]

Answer:

194 cents is the correct answer

Step-by-step explanation:

6.99 dollars is 41.94

7 0
3 years ago
Read 2 more answers
What is the sum of the interior angles of a convex 14-gon?
Ad libitum [116K]
I hope this helps you

6 0
3 years ago
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