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mamaluj [8]
3 years ago
15

In (x + 2) (3x + 1) = 3x^2 + ___ + 2, what should be the value of the middle term to complete the product?

Mathematics
1 answer:
zvonat [6]3 years ago
5 0

Answer:

middle term = 7x

Step-by-step explanation:

in order to find the middle term, we will expand the bracket on the left hand side and compare the expressions on both sides.

(x + 2) (3x + 1) = 3x² + ___ + 2

let the unknown expression = y

(x + 2) (3x + 1) = 3x² + y + 2

Expanding the bracket on the left hand side by multiplication:

x + 2( 3x + 1) = (x × 3x) + (x × 1) + (2 × 3x) + ( 2 × 1)

= 3x² + x + 6x + 2

= 3x² + 7x + 2

Comparing with the expression on the right hand side:

3x² + 7x + 2 =  3x² + y + 2

∴ y = 7x

middle term = 7x

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4 0
3 years ago
Solve for r. Show your work!<br> 12 =R– (34 – 2)<br><br> Helpppp!!
amm1812

Answer:

r = 44

Step-by-step explanation:

12 = r - (34 - 2)

Solve for r

___________

To solve, we need to isolate r.

Distribute the negative sign with the numbers within the parenthesis :

12 = r - 34 + 2

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4 0
3 years ago
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Let f(x) = (5)^x+1 .Evaluate f (-3) without using a calculator
rewona [7]

Answer:

f\left(-3\right)=\frac{126}{125}

Step-by-step explanation:

Given

f\left(x\right)\:=\:\left(5\right)^x+1

Putting x = -3 to find f(-3)

f\left(-3\right)\:=\:\left(5\right)^{\left(-3\right)}+1

as

5^{-3}=\frac{1}{125}       ∵  \mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

so

f\left(-3\right)=\frac{1}{125}+1

\mathrm{Convert\:element\:to\:fraction}:\quad \:1=\frac{1\cdot \:125}{125}

f\left(-3\right)=\frac{1\cdot \:\:125}{125}+\frac{1}{125}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

f\left(-3\right)=\frac{1\cdot \:\:125+1}{125}

f\left(-3\right)=\frac{126}{125}      ∵ 1\cdot \:125+1=126

Thus,

f\left(-3\right)=\frac{126}{125}

6 0
3 years ago
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