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RUDIKE [14]
4 years ago
10

2(x-11)+5-1 What is x

Mathematics
2 answers:
Radda [10]4 years ago
7 0

Answer:

2x-18

Step-by-step explanation:

2(x-11)+5-1\\

Expand

2\left(x-11\right):\quad 2x-22

Slot it into the equation

=2x-22+5-1

Add/subtract the numbers

-22+5-1=-18\\\\=2x-18

dangina [55]4 years ago
5 0

Answer:

9

Step-by-step explanation:

2(x-11) + 5-1

We can use the distributive property:

2x - 22 + 5 - 1

We can then simplify:

2x - 18

In this case, if 2x - 18 = 0, then 2x = 18 and x = 9.

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1/6 of 3 yards=______feet<br><br>I know I just ask one but u get 17 points so plz help I'm stuck
bulgar [2K]

In this question, first we have to see how many feets are in 1 yard .

And there are 3 feets in 1 yard.

So in 3 yards, there are 9 feets

Therefore we will get

\frac{1}{6}*3 yards= \frac{1}{6} * 9 feet = 1.5 feet

Therefore the answer of the given question is 1.5 feet.

3 0
4 years ago
There are 20 students on the schools student council. A special homecoming dance committee is to be formed by randomly selecting
babymother [125]
You have to take 20/7 and you get a answer of <span>2.85 as a decimal. or a mixed number of 2 17/20</span>
3 0
3 years ago
An insect is 3 inches long. What is the length to the nearest hundredth of a centimeter
densk [106]
1 inch = 2.54 cm
3 inches = 3 * 2.54 = 7.62 cm
6 0
3 years ago
Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k
Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

e=\frac{c}{a}

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

0.0444=\frac{c}{2,876,769,540 }

c=127728567.576

Minimum distance of Uranus from the sun is:

a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

6 0
3 years ago
50 POINTS!!! Word problem.
oksano4ka [1.4K]
Let d = # of dimes
Let q = # of quarters
45 = d + q
.1d + .25q = 8.40
Then we can use the substitution method and make it so that d = 45 - q.
Then plug in the value for d into the second equation.
.1(45-q)+.25(q)=8.40
4.5-.1q+.25q=8.40
4.5+.15q = 8.40
.15q = 3.90
q = 26
d = 45 - 26 = 19
26 quarters
19 dimes
4 0
3 years ago
Read 2 more answers
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