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Alexeev081 [22]
3 years ago
10

Sarah uses 9.1 pints of blue paint and white paint to paint her bedroom walls. 4 5 of this amount is blue paint, and the rest is

white paint. How many pints of white paint did she use to paint her bedroom walls?
Mathematics
1 answer:
bonufazy [111]3 years ago
4 0

Answer:Amount of white  paint used by Sarah =1.82 pints

Step-by-step explanation:

Amount of  blue paint and white paint to paint her bedroom walls.=9.1 pints

Amount of  blue paint used =4/5 x 9.1=7.28 pints

Amount of white  paint used=Amount of  blue paint and white paint to paint her bedroom walls -Amount of  blue paint used

=9.1-7.28=1.82 pints

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An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 25 ft.
faltersainse [42]

Answer:

y=-\frac{1}{100}(x-50)^2+25

the height of the arch 10 feet from the center is 24 feet

Step-by-step explanation:

An arch is in the shape of a parabola. It has a span of 100 feet, the vertex lies at the center 50 and the maximum height of 25 ft.

Vertex at (50,25)

vertex form of the equation is

y=a(x-h)^2+k, (h,k) is the center

y=a(x-50)^2+25

the parabola starts at (0,0) that is (x,y)

0=a(0-50)^2+25

subtract 25 from both sides

-25=a(-50)^2

-25=a(2500)

divide both sides by 2500

a=-\frac{1}{100}

y=-\frac{1}{100}(x-50)^2+25

the height of the arch 10 feet from the center.

center is at 50, 10 feet from the center so x=40 and x=60

y=-\frac{1}{100}(40-50)^2+25

y=24

the height of the arch 10 feet from the center is 24 feet

4 0
3 years ago
SIMPLIFY and SHOW WORK THANKS 3∙[ 9 – 2∙ (7 – 8)]
Zepler [3.9K]

Answer:

33

Step-by-step explanation:

3∙[ 9 – 2∙ (7 – 8)]

PEMDAS,

Parentheses, start from the inside out

3∙[ 9 – 2∙ (-1)]

3∙[ 9 +2]

3* 11

33

8 0
3 years ago
If f(x)=2x^2sqrt(x-2), complete the following statement f(6)
Sergio [31]
Hello : 
f(6) =2(6)² +<span>√(6-2) = 72 +2 =74</span>
8 0
3 years ago
If (x - h) = 4 and (y - k) = 3, what is the radius of the circle above?
I am Lyosha [343]
(x-h)²+(y-k)²=r² is the equation of a circle with radius of r


so if x-h=4 and y-k=3 then

4²+3²=r²
16+9=r²
25=r²
sqrt both sides
5=r
radius is 5 units
5 0
3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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