Simplify the product using the distributive property. (5h-3)(3h+7)

2 answers:

7 0

Use FOIL (first, outer, inner, last) :
(<span>5h-3)(3h+7)

</span>(5h)(3h)+7(5h)-3(3h)-3(7)
15h^2+35h-9h-21

D: 15h^2+26h-21

Hope this helps :)

UkoKoshka [18]3 years ago

5 0

Answer:

Option D. 15h² + 26h - 21

Step-by-step explanation:

We have to simplify the product by using distributive property.

(5h - 3)(3h +7) = 5h(3h + 7) - 3(3h + 7)

= 15h² + 35h - 9h -21

= 15h² + 26h -21

Which matches the option D.

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Line segment ST is dilated to create line segment S'T' using the dilation rule DQ,2. 25. Point Q is the center of dilation. Line

abruzzese [7]

The distance between points S' and S is x= 2.5 units.

<h2>Given that</h2>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ 2. 25.

Point Q is the center of dilation.

Line segment ST is dilated to create line segment S prime T prime.

The length of QT is 1. 2 and the length of QS is 2.

The length of SS prime is x and the length of TT prime is 1. 5.

<h3>We have to determine</h3>

What is x, the distance between points S' and S?

<h3>According to the question</h3>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ, 2.25.

Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5, SS' = x.

Since the line ST is dilated to S'T' with the center of dilation Q, the triangles STQ and S'T'Q must be similar.

We know that the corresponding sides of two similar triangles are proportional.

So, from ΔSTQ and ΔS'T'Q.

$\dfrac{SQ}{S'Q} = \dfrac{TQ}{T'Q}\\
\\
\dfrac{SQ}{SQ+SS} = \dfrac{TQ}{TQ+TT'}\\
\\ 
\dfrac{2}{2+x} = \dfrac{1.2}{1.2+1.5}\\\rm 
\\
\dfrac{2}{2+x} = \dfrac{1.2}{2.7}\\\\ \dfrac{2}{2+x} = \dfrac{12}{27}\\\\2(27) = (2+x) 12\\\\ 54 = 24 + 12x\\
\\
12x = 54-24\\
\\
12x=30\\
\\
x = \dfrac{30}{12}\\
\\
x = 2.5$