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hjlf
3 years ago
5

Can someone help me? Forgot how to do this, it’s been awhile

Mathematics
1 answer:
Aleks [24]3 years ago
7 0

Answer: y=5

Step-by-step explanation:

Similar basically means it's the same shape but bigger. Condiering how 16-9=7 which means each side is 7 bigger than the original shape. You subtract 7 from 12(side y) to get 5

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What is the solution to the following equation?
lions [1.4K]

Answer:

D. x = 7

Step-by-step explanation:

NOTE : <u><em>there should be an equal sign somewhere in the given expression.</em></u>

suppose the equation is the following:

3(x-4)-5 = x-3

………………………………………………………

3(x - 4) - 5 = x - 3

⇔ 3x - 12 - 5 = x - 3

⇔ 3x - 17 = x - 3

⇔ 3x - 17 + 17= x - 3 + 17

⇔ 3x = x + 14

⇔ 2x = 14

⇔ x = 14/2

⇔ x = 7

8 0
2 years ago
Which gives the correct values for points A, B, C, and D?
saveliy_v [14]

Answer:

<em>The correct answer is: A</em>

Step-by-step explanation:

<u>Points on a Numeric Line</u>

The numeric line shown in the figure has four points A, B, C, and D. It's clear that each marked division of the line has a value of 1/4 units and all four points are below the reference zero, thus all are negative.

Point A is clearly on the -2 mark, thus

A=-2

Point B is one mark above A, thus it's located at:

B=-2+\frac{1}{4}=-\frac{7}{4}=-1\frac{3}{4}

B=-1\frac{3}{4}

Point C is one mark below -1, thus:

C=-1-\frac{1}{4}=-1\frac{1}{4}

Finally, point D is one mark below 0:

D=-\frac{1}{4}

The correct answer is: A

5 0
3 years ago
If jimmy had had 9 apples and karl has 10 how mush togheter
sveticcg [70]
19? Lol........................
7 0
2 years ago
Use the quadratic formula to solve the equation.–x2 + 5x = 3
Karo-lina-s [1.5K]

Given the equation - x² + 5x = 3, which can be rewritten as:

- x² + 5x - 3 = 0

where a = -1, b = 5 and c = -3.

Quadratic formula:

\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}

Now, we just replace the values of a, b and c on the equation above.

\frac{-5\text{ }\pm\text{ }\sqrt[]{5^2\text{ - 4(-1)(3)}}}{2(-1)}

=

\frac{5}{2}\text{ }\pm\text{ }\frac{\sqrt[]{13}}{2}

4 0
1 year ago
7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and
Juliette [100K]

Answer:

A) JK and AB =

[tex] \boxed{( \theta) = \f{12}{( \theta)} } So, =》 \dfrac{10}{5} = \dfrac{1}{ \( \theta) } =》 \( \theta) = \dfrac{5}{2}[/tex]

B) AABC

[tex] \boxed{( \theta) = {1}{( \theta)} } So, 23}{-18=\\Ans } \boxed

C) AABM:

[tex] e \fbox{: } The value of k for which \sin(jm) = \cos(x) is \dfrac{\p}{2} ans[/tex]

----------

3 0
3 years ago
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