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Anastaziya [24]
4 years ago
6

PLZZZ HELP!!!! (WILL GIVE BRAINLIEST!!!!!!!)

Mathematics
1 answer:
tatiyna4 years ago
4 0
Y = -5x -5

To check, we can insert 0 for y and -1 for x

0 ? (-5X-1) - 5
0 ? 5 - 5
0 = 0

So yes, the equation is correct
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What is the value of x?<br> O 40<br> O 70<br> O 140<br> O 160
Semenov [28]

what is the question? if u give the question , I might help and edit this :)

7 0
3 years ago
A student wrote a photography manual which she plans to sell online for $ 13 per copy. Her total expenses were $ 55. Write a lin
finlep [7]

Answer:

y = 13x - 55

Step-by-step explanation:

So this is what we know

Selling price = $13 per copy

Expenses = $55

The formula for profit in this case is:

Profit = total selling price - expenses

In which y is the profit

The total selling price is 13x. x here is for the number of manuals she sells

The expenses are $55.

This then becomes y =13x - 55

HOPE THIS HELPED

5 0
3 years ago
Find the value of the expression 4a4 − 2b2 + 40 when a = 2 and b = 7
vovangra [49]

Answer:

74

Step-by-step explanation:

4(2) raise to power 4 - 2(7) raise to power 2 +40

4(16) - 2(49) +40

64-98+40

64- 138

74

4 0
3 years ago
Solve the equation and this is an assignment btw​
lilavasa [31]

Answer:

z=6 and z=-6

Step-by-step explanation:

7z^2=252

Divide both sides by 7

\frac{7z^2}{7} =\frac{252}{7} \\z^2=36

Find the square root

z=±√36

z=±6

Therefore, z=6 and z=-6.

I hope this helps!

4 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
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