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maria [59]
3 years ago
13

Seventeen minus 6 times 10 divided two plus twelve

Mathematics
2 answers:
Furkat [3]3 years ago
6 0
17-6×10÷2+12
Apply bodmas
=17-6×5+12
=17-30+12
=17-18
=-1
anyanavicka [17]3 years ago
6 0

Answer:

-1

Step-by-step explanation:

Using PEMDAS

17-6*10/2+12

17-30+12

equals -1

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The rectangular room shown has a length of 20 ft, width of 48 ft, and height of 10 ft. Use the Pythagorean Theorem to find lengt
Vlad [161]
For the diagonal of the floor:
20^2 + 48^2= c^2
400+2304= c^2
c=52
For the diagonal of the cube:
52^2 + 10^2 = c^2
2704+100=c^2
c=52.95
Rounded= 53
Tell me if this helps!!!
4 0
3 years ago
Read 2 more answers
The picture is my question.
Dominik [7]
The second one because domain refers to x, and since at (-3,-4) the point is hollow, the sign should be < without the line under it
5 0
3 years ago
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Trey believes there is a correlation between the number of texts sent during class and GPA. He collected data and found that the
vichka [17]

Answer: First Option

The slope is −0.4. Starting at 3.8, the GPA will decrease by 0.4 for every text sent in class.

Step-by-step explanation:

The equation of a line has the following form

y=mx+b

Where m is the slope of the line and b is the intercept with the axis of the y

In this case, the equation is:

y = 3.8 - 0.4x\\\\y=-0.4x+3.8

Observe that

m = -0.4\\b = 3.8

So if x represents the number of texts sent during classes, this means that x\geq 0

When x increases by one unit then y decreases -0.4. For example when x=1 then y=3.8-0.4=3.4

Then the greatest value it can take the GPA is 3.8. Therefore we can say that the function starts at y = 3.8.

The answer is the first option

5 0
3 years ago
Read 2 more answers
Identify the equation that represents the data in the table
ella [17]

Answer:

i would say A

Step-by-step explanation:

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%20%3D%200" id="TexFormula1" title=" {x}^{2} + 2x = 0" alt=
Alex

Answer:

\textbf{Hello!}

\Longrightarrow2^2+2z=0

\Longrightarrow x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \:0}}{2\cdot \:1}

\Longrightarrow \sqrt{2^2-4\cdot \:1\cdot \:0}

\Longrightarrow =\sqrt{2^2-0}

\Longrightarrow =\sqrt{2^2}

\Longrightarrow=2

\Longrightarrow x_{1,\:2}=\frac{-2\pm \:2}{2\cdot \:1}

\Longrightarrow x_1=\frac{-2+2}{2\cdot \:1},\:x_2=\frac{-2-2}{2\cdot \:1}

\Longrightarrow\frac{-2+2}{2\cdot \:1}

\Longrightarrow =\frac{0}{2\cdot \:1}

\Longrightarrow =\frac{0}{2}

=0

\Longrightarrow\frac{-2-2}{2\cdot \:1}

\Longrightarrow =\frac{-4}{2\cdot \:1}

\Longrightarrow =\frac{-4}{2}

\Longrightarrow =-\frac{4}{2}

=-2

x=0,\:x=-2\Longleftarrow

\underline{HOPE ~IT~HELPS}

5 0
3 years ago
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