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Anit [1.1K]
4 years ago
12

Which choices are solutions to the following equation?

Mathematics
2 answers:
podryga [215]4 years ago
3 0
You can see solution in the picture, hope that can help you :-)

Vilka [71]4 years ago
3 0

Answer:

B. x = 1

C. x = 6

Step-by-step explanation:

The equation

x^2 - 7x = -24/4

is the same as

x^2 - 7x = -6

We will verify which of the options meets this.

  • A. x=12

is x=12, the equation will be

(12)^2 - 7(12)

144 - 84

60

since is not equal to -6 this is not a solution.

  • B. x=1

is x=1, the equation will be

(1)^2 - 7(1)

1 - 7

-6

since is equal to -6 <u>this is a solution.</u>

  • C. x=6

is x=6, the equation will be

(6)^2 - 7(6)

36 - 42

-6

since is equal to -6<u> this is a solution.</u>

  • D. x=2

is x=2, the equation will be

(2)^2 - 7(2)

4 - 14

-10

since is not equal to -6 this is not a solution.

the solutions are: x=1 and x=6 which are options B and C

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-3v+24 would be your answer
5 0
3 years ago
On Friday, you raked leaves for 4 neighbors, on saturay you raked leaves for 5 neighbors, and on Sunday you raked leaves for 3 n
Jet001 [13]

Answer:

The amount paid by each house is $11.25

Step-by-step explanation:

Let the amount paid by each house is $K

Number of neighbors you worked for on Friday = 4

So the amount paid by all of them = 4K

Number of neighbors you worked for on Saturday = 5,

similarly amount paid = 5K

Number of neighbors you worked for on Sunday = 3,

here amount paid = 3K

Also, in total the amount paid = $135

So, Amount paid on {Friday + Saturday +Sunday} = Total Amount

or, 5K + 4K + 3K = $135

or, 12K = 135

⇒K =135/12 = $11.25

So, the amount paid by each house is $11.25

6 0
3 years ago
Find all values of c such that 3^2c+1=28*3^c-9. If you find more than one value of c, then list your values in increasing order​
avanturin [10]

3^{2c} + 1 = 28\times3^c - 9 \implies 3^{2c} - 28\times3^c = -10

Complete the square on the left side:

3^{2c} - 28\times3^c = \left(3^{2c}-28\times3^c+14^2\right)-14^2 = \left(3^c-14\right)^2 - 196

Then the equation becomes

\left(3^c-14\right)^2 - 196 = -10 \\\\ \left(3^c-14\right)^2 = 186 \\\\ 3^c - 14 = \pm\sqrt{186} \\\\ 3^c = 14\pm\sqrt{186}

Both 14 + √186 and 14 - √186 are positive numbers, so we can take the logarithm (base 3) of both sides without issue:

\log_3\left(3^c) = c = \log_3\left(14\pm\sqrt{186}\right)

Then in increasing order, the solutions are

<em>c</em> = log₃(14 - √186), <em>c</em> = log₃(14 + √186)

3 0
3 years ago
Help i need this..........................
Maurinko [17]

Answer:

wish i could help sorry have a great day

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Ten less than eight times the number of marbles is greater than or equal to the sum of six times the number of marbles and fourt
tia_tia [17]

Let, the number = x

8x-10 ≥ 6x+14

Subtracting 6x form both sides,

2x-10 ≥ 14

Adding 10 to both sides,

2x ≥ 24

Dividing by 2,

x ≥ 12


3 0
3 years ago
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