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3 years ago

7

Brandon bought a 5-gallon container of paint to paint his house. After he finished painting, he had 2 quarts of paint left over.

How many quarts of paint did Brandon use?

1 answer:

UNO [17]3 years ago

5 0

Answer:

18 Quarts

Step-by-step explanation:

there are 4 quarts per gallon

I multiplied 4 x 5 = 20 then subtracted the 2 quarts left which comes out to 18 quarts used.

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Going into the final exam, which will count as two tests, Brooke has test scores of 75, 82, 70, 65, and 90. What score does Broo

dusya [7]

Brooke needs to get a 98 on the final to get an average of 80.
So you know that the total of the five given score is 382 by adding all the scores.
Now you multiply the average, 80 by 6 for the total of six tests including the final which is 480
Then you subtract 480-382 to find your answer.

4 0

3 years ago

Using the method of completing the square, put each circle into the form

tatiyna

Answer:

Standard form: $(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Center: $(\frac{1}{2}, 0)$

Radius: $r =\sqrt{15}$

Step-by-step explanation:

The equation of a circle in the standard form is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Where the point (h, k) is the center of the circle

To transform this equation $4x^{2} -4x + 4y^{2} - 59 = 0$ this equation in the standard form we use the method of square.

First, we group similar variables

$(4x^{2} -4x) + (4y^{2}) - 59 = 0$

Divide both sides of equality by 4

$(x^{2} -x) + (y^{2}) - 14.75 = 0$

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

$b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}$

Now add $(\frac{b}{2})^2$ on both sides of the equality

$(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})$

Factor the expression and simplify the independent terms

$(x-\frac{1}{2})^2 + (y^{2}) = 15$

$(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Then

$h =\frac{1}{2}\\\\k=0$

and the center is $(\frac{1}{2}, 0)$

radius $r =\sqrt{15}$

3 0

3 years ago

Plz Help! I will mark brainliest

blondinia [14]

Answer:

the answer is C

(5+3i)-(2+9i) = 3-6i

5 0

3 years ago

Read 2 more answers

Rampur Sarpanch requested one of his villager to donate a 6m wide land adjusted to his 132.8m long side of his right triangular

Furkat [3]

Answer:

The answer is 3030.72 m³

Step-by-step explanation:

Please refer to attached file for graphical representation.

Formulas used:

Area = (a*b)/2

Trigonometry: Tan (t) = opposite/adjacent

Assume:

Villager plot has 3 sides: hypotenuse=C, adjacent=A, opposite=B

Plot given has 3 sides: hypotenuse=z, adjacent=y, opposite=x

Angle common between them is t.

<u>Step 1</u>:

For t:

Tan (t) = a/b

Tan (t) = 123/50

Tan (t) = 2.46

<u>Step 2:</u>

For y:

Tan (t) = y/x

2.46 = y/6

y = 2.46*6

y = 14.76

<u>Step 3:</u>

Villagers old plot area: B = (a*b)/2 = (123*50)/2

B = 3075 m³

<u>Step 4:</u>

Plot donated area: C = (x*y)/2 = (6*14.76)/2

C = 44.28 m³

<u>Step 5:</u>

Villager remaining plot:

A = B-C

A= 3075-44.28

A = 3030.72 m³

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3 years ago

Help 12 minutes to turn in

NARA [144]

Shows how far day valued asking questions

6 0

3 years ago