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blondinia [14]
3 years ago
11

Evaluate 5x + 3 when x=6

Mathematics
1 answer:
charle [14.2K]3 years ago
7 0

Answer:

33

Step-by-step explanation:

5(6) + 3

30 + 3

= 33

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What is the value of 7 ^−3 ^ −1 for = −2 and = 4?
andrey2020 [161]

Answer:

d. −7/32

Step-by-step explanation:

The expression that we have to evaluate in this problem is:

7x^{-3}y^{-1}

We have to evaluate this expression for:

x = -2

y = 4

We start by rewriting the expression by rewriting it using the following:

a^{-n}=\frac{1}{a^n}

So the expression can be rewritten as

7x^{-3}y^{-1}=\frac{7}{x^3 y}

Now we observe that:

(-2)^3=(-2)(-2)(-2)=-8

Therefore, by substituting x = -2 and y = 4 into the expression, we find:

\frac{7}{(-2)^3\cdot 4}=\frac{7}{-8\cdot 4}=-\frac{7}{32}

8 0
3 years ago
52 /72
Stolb23 [73]
52/72

13/18 is your answer. Therefore, C.
8 0
3 years ago
Please explain and show work ( I WILL MARK BRAINLIEST)
kvasek [131]

Answer:

-2

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
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zhannawk [14.2K]
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3 years ago
Read 2 more answers
Determine tan(t) if cos(t)= -3/5 and sin(t) >0
kupik [55]
\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\
cos(t)=-\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}

now, the hypotenuse is just a radius unit, so, is never negative, so, the fraction is negative because the numerator is negative, that is, the adjacent side is -3.

now, let's use the pythagorean theorem to find the opposite side.

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{5^2-(-3)^2}=b\implies \pm\sqrt{25-9}=b\implies \pm 4=b

ok... so, which is it? the +/-?   well, we also know that <span>sin(t) >0, namely that the sine of the angle is positive, so, then  is +4 then.

</span>\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(t)=\cfrac{4}{-3}\implies \boxed{tan(t)=-\cfrac{4}{3}}<span>
</span>
4 0
3 years ago
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