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Sidana [21]
3 years ago
10

I need help (last question)

Mathematics
1 answer:
svet-max [94.6K]3 years ago
7 0
What’s the last question? please attach a photo
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Jamal wrote the inequality x/16 <6 . Which situation best represented by this inequality Helpppppp
Wittaler [7]

Given problem;

   Inequality equation;

               \frac{x}{16}   <   6

Before we solve, let us first translate this problem.

It simply states that for what value(s) of x will the expression be less than 6.

To find this value, we can simply carry out the normal mathematical simplification.

Simply multiply both sides by 16 to reduce the fraction;

      16  x (\frac{x}{16})   <  6 x 16

         

 On the left hand side, 16 will cancel out;

                   x <  96

Any value for which x is less than 96 will make the solution of this problem less than 6.

For example, 95;

                       \frac{95}{16}  = 5.93

This value is less than 6

4 0
3 years ago
What is the missing side length in the triangle?<br><br> plsss someone help!!
UkoKoshka [18]

Answer:

The measure of the missing side length = 45

Step-by-step explanation:

Pythagorean Theorem states: A² + B² = C²

28² + B² = 53²

784 + B² = 2809

B² = 2809 - 784

B² = 2025

B = \sqrt{2025}

B = 45

Therefore the measure of the missing side length = 45

Hope this helps!

6 0
3 years ago
Read 2 more answers
Next three terms in (1024, -128, 16...) you can write them as fraction if needed
Tcecarenko [31]

Answer:

-2 1/4 - 1/32

Step-by-step explanation:

16x -1/8 = -2 do the same for all

8 0
3 years ago
A suit that sells for $300 in on sale of $240. Find the discount rate.
Gemiola [76]

Answer:

20% discount

Step-by-step explanation:

300 - 240 = 60

60 / 300 = 0.20

0.20 × 100 = 20

3 0
3 years ago
The point (2, 1) is a solution to which of the following systems of equations?
Snezhnost [94]

\huge\boxed{\left \{ {{x+y=3} \atop {3x-y=5}} \right.}

We can solve this problem by testing the solution against each system of equations. We'll do this by substituting the point into each equation for each system.

<h2>System 1 - \Large\textbf{X}</h2>

\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}2x+y&=5\\2(2)+1&=5\\4+1&=5\\5&=5\\&\checkmark\end{aligned}&\begin{aligned}-2x+y&=2\\-2(2)+1&=2\\-4+1&=2\\-3&=2\\&\text{X}\end{aligned}\end{array}

<h2>System 2 - \Large\textbf{X}</h2>

\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-2x-y&=-5\\-2(2)-1&=-5\\-4-1&=-5\\-5&=-5\\&\checkmark\end{aligned}&\begin{aligned}2x-y&=2\\2(2)-1&=2\\4-1&=2\\3&=2\\&\text{X}\end{aligned}\end{array}

<h2>System 3 - \Large\textbf{X}</h2>

\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-x+y&=3\\-2+1&=3\\-1&=3\\&\text{X}\end{aligned}&\begin{aligned}-3x+y&=-5\\-3(2)+1&=-5\\-6+1&=-5\\-5&=-5\\&\checkmark\end{aligned}\end{array}

<h2>System 4 - \Large\checkmark</h2>

\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}x+y&=3\\2+1&=3\\3&=3\\&\checkmark\end{aligned}&\begin{aligned}3x-y&=5\\3(2)-1&=5\\6-1&=5\\5&=5\\&\checkmark\end{aligned}\end{array}

Since both equations in the final system are true when solved with (2, 1), the answer is the last system.

3 0
3 years ago
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