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4 years ago

draw and label a parallelogram with a base twice as long as the height and an area less than 60 square inches. find the area

1 answer:

3 0

Answer: 5 cm by 10 cm gives an area of 50 cm^2

To do this draw a parallelogram and start by labeling the base as 10 cm. If 10 cm is the base, then the height needs to be 5 cm.

To find the area, multiply the base times height.

This will give you: 5 x 10 = 50 which is less than 60.

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3x − x − 1 = 15<br> What is x

azamat

Answer: x=8

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

3x−x−1=15

3x+−x+−1=15

(3x+−x)+(−1)=15(Combine Like Terms)

2x+−1=15

2x−1=15

Step 2: Add 1 to both sides.

2x−1+1=15+1

2x=16

Step 3: Divide both sides by 2.

2x /2 = 16 /2

x=8

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Divide 9x ^ 3 + 18x ^ 2 - 13x + 5 by 3x - 1 using division and write the division in the form P = DQ + R

lana66690 [7]

Answer:

$\frac{9x^3\:+\:18x\:^2\:-\:13x\:+\:5}{3x-1}=9x^3+18x^2-13x+5$

Step-by-step explanation:

DIVISION ALGORITHM: If $p(x)$ and $d(x)\neq 0$ are polynomials, and the degree of $d(x)$ is less than or equal to the degree of $f(x)$ , then there exist unique polynomials $q(x)$ and $r(x)$ , so that

$\frac{p(x)}{d(x)} =q(x)+\frac{r(x)}{d(x)}$

and so that the degree of $r(x)$ is less than the degree of $d(x)$ .

To find $\frac{9 x^{3} + 18 x^{2} - 13 x + 5}{3 x - 1}$ you must:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{9x^3}{3x}=3x^2$

$\mathrm{Quotient}=3x^2$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}3x^2:\:9x^3-3x^2\\\mathrm{Subtract\:}9x^3-3x^2\mathrm{\:from\:}9x^3+18x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\$

$\mathrm{Remainder}=21x^2-13x+5$

Therefore,

$\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+\frac{21x^2-13x+5}{3x-1}$

$\mathrm{Divide}\:\frac{21x^2-13x+5}{3x-1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{21x^2}{3x}=7x\\\\\mathrm{Quotient}=7x$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}7x:\:21x^2-7x\\\mathrm{Subtract\:}21x^2-7x\mathrm{\:from\:}21x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-6x+5$

Therefore,

$\frac{21x^2-13x+5}{3x-1}=7x+\frac{-6x+5}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x+\frac{-6x+5}{3x-1}$

$\mathrm{Divide}\:\frac{-6x+5}{3x-1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{-6x}{3x}=-2\\\\\mathrm{Quotient}=-2$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}-2:\:-6x+2\\\mathrm{Subtract\:}-6x+2\mathrm{\:from\:}-6x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=3$

Therefore,

$\frac{-6x+5}{3x-1}=-2+\frac{3}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x-2+\frac{3}{3x-1}$

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4 years ago