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3 years ago

What is the product of (3 + x) and (2x – 5)

Mathematics

1 answer:

lozanna [386]3 years ago

5 0

Answer:

2x² + x - 15

Step-by-step explanation:

(3 + x)(2x - 5)

Each term in the second factor is multiplied by each term in the first factor, that is

3(2x - 5) + x(2x - 5) ← distribute both parenthesis

= 6x - 15 + 2x² - 5x ← collect like terms

= 2x² + x - 15

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Answer:

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Step-by-step explanation:

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3 years ago

Round 62.276 to the nearest hundredth

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3 years ago

Which equation can be simplified to find the inverse of y = 5x² + 10?

notsponge [240]

Answer:

See below

Step-by-step explanation:

Here is how to start....you didn't include the equations in question, so I do not know what the answer may be

solve for x

y-10 = 5 x^2 divide thru by 5

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2 years ago

Consider the trinomial below x^2-20x+91 The factors of the given trinomial are

VashaNatasha [74]

x² - 20x + 91
x -13
x -7

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hope this helps

5 0

3 years ago

Read 2 more answers

In △ABC, point D∈ AC with AD:DC=4:3, point E∈ BC so that BE:EC=1:5. If ACDE=5 in2, find ABDC, AABD, and AABC.

Kaylis [27]

Answer:

ar ΔBDC=6 $in^2$

ar ΔABD=8 $in^2$

ar ΔABC=14 $in^2$

Explanation:

Please take a look of attach figure.

We are given area of triangle ar ΔCDE=5 $in^2$

ar ΔCDE= $\frac{1}{2}\times DN\times EC$ ------------(1)

ar ΔBDE= $\frac{1}{2}\times DN\times BE$ ------------(2)

Divide eq(2) by eq(1) and we get,

$\frac{ar\ ΔBDE}{ar\ ΔCDE}=\frac{BE}{EC}$

$\frac{ar\ ΔBDE}{5}=\frac{1}{5}$

So, ar ΔBDE=1 $in^2$

ar ΔBDC=ar ΔBDE+ ar ΔCDE

ar ΔBDC=1+5 = 6 $in^2$

Similarly,

$\frac{ar\ ΔABD}{ar\ ΔBDC}=\frac{AD}{DC}$

$\frac{ar\ ΔABD}{6}=\frac{4}{3}$

ar ΔABD=8 $in^2$

ar ΔABC=ar ΔABD + ar ΔBDC

ar ΔABC=8+6 = 14 $in^2$

6 0

3 years ago