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damaskus [11]
3 years ago
8

What is the value of x in the equation 2(x – 3) + 9 = 3(x + 1) + x? x =

Mathematics
2 answers:
dimulka [17.4K]3 years ago
8 0

Answer:

im not sure

Step-by-step explanation:

Mademuasel [1]3 years ago
7 0
It means the same:

2x - 6 + 9 = 3x + 3 + x
2x +3 = 4x + 3
2x - 4x = 3 - 3
-2x = 0
x = 0/(-2) = 0

Answer:
x = 0
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Help please and explain
tresset_1 [31]

Answer:

30 ft

Step-by-step explanation:

We can use ratios

3              6

------ = ------------

15            x

Using cross products

3x = 6*15

Divide each side by 3

3x/3 = 6*15/3

x = 30

5 0
3 years ago
Read 2 more answers
Need help with this answer
ioda

Answer:

(2,2) is your answer :)

Step-by-step explanation:

3 0
3 years ago
If the circumference is 31.4, what are the radius and the area?
jekas [21]

Answer:

radius = 5 units

area of a circle = 15.7units²

Step-by-step explanation:

given that the circumference of a circle = 31.4

radius = ?

recall that the formulae for finding the circumference of a circle  =  2πr

circumference of a circle  = 2 x 3.14 x r

31. 4 = 6.28r

divide both sides by the coefficient of r which is 6.28

31.4/6.28 = 6.28r/6.28

5 = r

radius = 5 units

therefore the radius of the circle whose circumference is 31.4 units is evaluated to be 5units

to find the area of the circle

recall the formula = πr²

area of a circle = 3.14 x 5²

area of a circle = 3.14 x 25

area of a circle = 15.7units²

7 0
3 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
At the start of the year, the balance in a married couple's account is $1,750. They decide to each deposit $35 into the account
ANTONII [103]
After 1 year, the couple will have a total of $2170.00 into their account.
8 0
3 years ago
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