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Oksi-84 [34.3K]
3 years ago
8

Q # 1 please solve the equation

Mathematics
2 answers:
zepelin [54]3 years ago
6 0
The answer is 60 The answer is 60.
Bogdan [553]3 years ago
5 0
This is an exam! You are not supposed too cheat! >:O
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Alexander made of rectangular quilt that measured 3 1/4 feet in length by 2 3/4 feet and width to find the area multiply the len
Oksana_A [137]
\text {Area = } 3 \dfrac{1}{4}  \times 2 \dfrac{3}{4}

\text {Area = } \dfrac{13}{4}  \times \dfrac{11}{4}

\text {Area = } \dfrac{143}{16}

\text {Area = } 8 \dfrac{15}{16} \text { ft}^2
5 0
3 years ago
What is the slope of the line that passes through the points (-10,8)and (-15,-7)
balandron [24]

Answer:

m = 3

Step-by-step explanation:

Use the slope formula to find the slope  m .

5 0
3 years ago
Read 2 more answers
The slope of the line below is -5. Which of the following is the point slope form of the line (2,-7)
tatyana61 [14]

Plugging in (2,-7) into

y=-5x+b, we get that b=3.

Therefore, the equation is y=-5x+3

7 0
3 years ago
If a polynomial function f(x) has roots –8, 1, and 6i, what must also be a root of f(x)? A. –6
anyanavicka [17]

Answer:

-6i

Step-by-step explanation:

Complex roots always come in pairs, and those pairs are made up of a positive and a negative version. If 6i is a root, then its negative value, -6i, is also a root.

If you want to know the reasoning, it's along these lines: to even get a complex/imaginary root, we take the square root of a negative value. When you take the square root of any value, your answer is always "plus or minus" whatever the value is. The same thing holds for complex roots. In this case, the polynomial function likely factored to f(x) = (x+8)(x-1)(x^2+36). To solve that equation, you set every factor equal to zero and solve for the x's.

x + 8 = 0

x = -8

x - 1 = 0

x = 1

x^2 + 36 = 0

x^2 = -36 ... take the square root of both sides to get x alone

x = √-36 ... square root of an imaginary number produces the usual square root and an "i"

x = ±6i

6 0
3 years ago
Read 2 more answers
Consider the surface z = x 2+4y 2+1. Suppose you are walking on this surface directly above a curve C in the xy-plane, where the
Bess [88]

With x(t)=\cos t and y(t)=\sin t, we have

z(t)=x(t)^2+4y(t)^2+1

z(t)=\cos^2t+4\sin^2t+1

z(t)=3\sin^2t+2

Then z(t) has critical points where

\dfrac{\mathrm dz}{\mathrm dt}=6\sin t\cos t=3\sin2t=0

\implies\sin2t=0\implies2t=n\pi\implies t=\dfrac{n\pi}2

where n is any integer.

z(t) is increasing wherever \sin2t>0, which happens for

\dfrac{2n\pi}2

\boxed{n\pi

5 0
3 years ago
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