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Aleonysh [2.5K]
3 years ago
14

Pls help thx luv y’a

Mathematics
1 answer:
garik1379 [7]3 years ago
3 0

Answer:

8.39958×10(raised to power of eight)

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Sarah bought pencils and markers for school. She bought 3 times as many pencils as markers.
Serhud [2]

Answer:

Sarah bought 30 pencils

Step-by-step explanation:

1. Let define p, as the number of pencils bought, and m, as the number of markers bought.

2.

0.15p+0.30m=7.50

p=3m

3. Substitute p in the first equation to 3m since p=3m as defined.

0.15(3m) + 0.30m=7.50\\0.45m+0.30m=7.50\\0.75m=7.50\\m=10

plug m into equation p=3m to find the number of pencils bought

p=3m=3(10)=30

7 0
3 years ago
Please help.
S_A_V [24]
3. tan(α) = height/distance
distance = height/tan(α)
distance = (648 m)/tan(13°) ≈ 1399.5 m

6. when y = |x| is translated 6 units down, 6 is subtracted from the y-value:
.. y = |x| -6

10. In point-slope form, the line through point (h, k) with slope m is
.. y -k = m(x -h)
For (h, k) = (10, -9) and m = -2, you have
.. y +9 = -2(x -10)
6 0
3 years ago
Which is the solution to the inequality <br><img src="https://tex.z-dn.net/?f=7x%20-%205%20%5Cgeqslant%20x%20%2B%201" id="TexFor
earnstyle [38]
X
≥
1
, or, in the interval form,
x
∈
[
1
,
∞
)


5 0
3 years ago
PLEASE HELP! <br> Will try to give the brainliest!
dusya [7]

Answer:

L: (15+6x) W: 2

L: (10+4x) W: 3

L: (5+2x) W: 6

Step-by-step explanation:

We don't have any numbers here to do calculations, so probably this is just an exercise in FACTORING. Also to practice that area is length times width.

The area given,

30 + 12x

can be factored several ways to be able to fill in the table.

This kind of factoring is like "un-distributive property" where a common factor can be factored out of the expression. Usually its helpful to factor out the greatest common factor (GCF) but in this case any common factor will do.

Also, unless there are more directions, either factor may be the length or the width.

2(15+6x)

3(10+4x)

6(5+2x)

6 0
3 years ago
Find the area of the rhombus
MAXImum [283]
<h3> Solution :-</h3>

Area of rhombus :-

\bf  \star \:  \:  \boxed{ \bf  \dfrac{d_{1}d_{2}}{2} }

\sf \longrightarrow d_{1} = 4 \sqrt{3}  + 4 \sqrt{3} = 8 \sqrt{3}

\sf \longrightarrow d_{2}  = 4 + 4 = 8

Area of rhombus :-

:  \implies \sf  \dfrac{8 \sqrt{3} \times 8 }{2}

:   \implies \sf  8 \sqrt{3}  \times 4

:  \implies \bf 32 \sqrt{3}  \:  {m}^{2}

6 0
3 years ago
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