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Contact [7]
4 years ago
13

NEED HELP

Mathematics
1 answer:
Airida [17]4 years ago
7 0
The numbers are 11 and 6.

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Two storage sheds are to have the same area. One is square and one is rectangular. The rectangular shed is 4 meters wide and 9 m
aliina [53]
Area is calculated by taking the length times the width.
A=lxw

one shed is 9 x 4 =36 square feet.
the other shed will be the same but have a four sides equal.
a shed with side of 6 x 6 will have a area of 36 square feet.

one side of the square shed would be 6 feet.

brainliest is always appreciated..
6 0
3 years ago
F(x)-6(x-1)+9 evaluate f(3)
Lilit [14]

Answer:

fx-6x+15

Step-by-step explanation:

fx-(6x-6)+9

fx-6x+6+9

fx-6x+(6+9)

fx-6x+15

5 0
3 years ago
Fatuma invests a total of $14,000 in two accounts. The first account earned an annual interest rate of 15% and the second accoun
solong [7]

Answer: that’s 5

Step-by-step explanation:

3 0
3 years ago
Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution
Blababa [14]

Answer:

If you mean only one rational solution, the answer is

k_1 = 8, k_2 = -8

If you mean at least 1 rational solution, the answer is

k\in (-\infty, -8]\cup[8, \infty)

Step-by-step explanation:

4x^2 + kx + 4 = 0

Let's calculate the discriminant.

\Delta = b^2 - 4ac

\Delta = k^2 -4 \cdot 4 \cdot 4

\Delta = k^2 -64

Now, remember that:

\text{If } \Delta > 0 : \text{2 Real solutions}

\text{If } \Delta = 0 : \text{1 Real solution}

\text{If } \Delta < 0 : \text{No Real solution}

Therefore, I will just consider the first two cases.

k^2 - 64 > 0

and

k^2 -64 = 0

\boxed{\text{For } k^2 - 64 > 0}

k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64}   \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)

\boxed{\text{For } k^2 - 64 = 0}

k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8

7 0
3 years ago
Denzel is comparing web sites for downloading music.one charge a$5membership fee plus $0.5 per song .another charges $1.00 per s
MA_775_DIABLO [31]
.5x+$5=$1x
x=11
.5*11+5=11
1*11=11
4 0
4 years ago
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