All categories

3 years ago

Simplify: this please

Mathematics

1 answer:

joja [24]3 years ago

4 0

Answer:

$16\sqrt{3}$ , so therefore:

none of the above

You might be interested in

How many gallons did both pumps dispense

FrozenT [24]

What are the answer options

3 0

2 years ago

the sum of −2x2 + x + 31 and 3x2 + 7x − 8 can be written in the form ax2 + bx + c, where a, b, and c are constants. What is the

Kitty [74]

Answer:

$a+b+c=32$

Step-by-step explanation:

we have

$(-2x^{2} +x+31)+(3x^{2}+7x-8)$

Adds the terms

Group terms that contain the same variable

$(-2x^{2}+3x^{2})+(x+7x)+(31-8)$

Combine like terms

$(x^{2})+(8x)+(23)$

where

$a=1,b=8,c=23$

therefore

$a+b+c=1+8+23=32$

8 0

2 years ago

Karl's car went 550 miles on 33 gallons of gas. Which equation does NOT give the correct unit rate, x, in miles per gallon for K

jekas [21]

Answer:

x = 550 ÷ 33

what are the answer choises

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Enjoy the points :D eeeeeeeeeeeeee

balandron [24]

The answer is 4/8 because half of 16 is 8 and half of 8 is 4, therefore they are equivalent

3 0

2 years ago

Read 2 more answers

For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

professor190 [17]

The true statements are:

Tyson's expression is not equivalent to the original expression
The equivalent expression is: $\frac{1}{2}g- \frac 1{5}$

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}$

Collect like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}$

Evaluate the like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}$

Tyler's equivalent expression is given as:

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}$

Collect like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}$

Evaluate the like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g$

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0

1 year ago