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AURORKA [14]
3 years ago
12

How do you solve -2=u-93÷-3

Mathematics
2 answers:
HACTEHA [7]3 years ago
5 0
Ria is that you ...........
viktelen [127]3 years ago
4 0

Answer:

u=91

Step-by-step explanation:

-2=u-93÷-3

multiply both sides by 3

3(-2)=(u)3+(-93÷-3)3

-6=3u-279

add 279 to both sides

-6=3u-279

+279. +279

273=3u

divide by 3

273÷3=3u÷3

91=u

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4712.39

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3 years ago
There` a new question for 11
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Answer:

x=3 the angle is 48

Step-by-step explanation:

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8 0
3 years ago
- Does the graph below represent a function? How do you know? please answer! ​
andrezito [222]

Answer:

No, the graph does not represent a function because it does not pass the vertical line test.

<h2>What is a function?</h2>

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The yellow vertical line touches two points, which is more than one. So, the graph does not pass the test. This graph is does not represent a function.

5 0
2 years ago
a person kicks a ball from the ground into the air with an initial upward velocity of 8 feet per second. what is the maximum hei
KengaRu [80]
Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

\bf ~~~~~~\textit{initial velocity}&#10;\\\\&#10;\begin{array}{llll}&#10;~~~~~~\textit{in feet}&#10;\\\\&#10;h(t) = -16t^2+v_ot+h_o&#10;\end{array} &#10;\quad &#10;\begin{cases}&#10;v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\&#10;h_o=\stackrel{0}{\textit{initial height of the object}}\\\\&#10;h=\stackrel{}{\textit{height of the object at "t" seconds}}&#10;\end{cases}&#10;\\\\\\&#10;h(t)=-16t^2+8t+0

now let's find the y-coordinate of its vertex

\bf \textit{vertex of a vertical parabola, using coefficients}&#10;\\\\&#10;h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}&#10;\qquad \qquad &#10;\left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right)&#10;\\\\\\&#10;\left(\qquad ,~~0-\cfrac{8^2}{4(-16)}  \right)\implies (\quad ,~0+1)\implies (\quad ,~1)

when will it hit the ground?

\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0&#10;\\\\\\&#10;8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}&#10;\\\\\\&#10;2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it  when it came back down}}{t=\cfrac{1}{2}}

3 0
3 years ago
Help me pleasee Jesus (10 points )
Bad White [126]

Answer:

Pretty sure it's 29

Step-by-step explanation:

The triangle has to be 180 degrees so 15+36=51 180-51=29

4 0
3 years ago
Read 2 more answers
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