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3 years ago

What is the number that represents C in 4c + 16 = 35?

Mathematics

2 answers:

Elan Coil [88]3 years ago

6 0

C=15 is the correct answer.

raketka [301]3 years ago

4 0

15, if you ad 4 and 16 (20) and subtract 35 you should get 15. You welcome

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Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Luba_88 [7]

Answer:

$B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15$

$D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3$

$H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4$

$L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9$

Step-by-step explanation:

We need to match the slope of the function with the slope of the lines connecting the two points given. The slope of the lines are as follows:

$B(5,2) \:\: C(7,-5)=\frac{2--5}{5-7} =-3.5$

$D(11,6)\:\: E(5,9)=\frac{6-9}{11-5} =-0.5$

$H(4,4)\:\: I(8,9)=\frac{4-9}{4-8} =1.25$

$L(5,-7)\:\: M(4,-12)=\frac{-7--12}{5-4} =5$

$F(-7,12)\:\: G(3,-8)=\frac{12--8}{-7-3}=-2$

$J(7,2)\:\:K(-9,8) =\frac{2-8}{7--9} =-0.375$

Now,

the slope of the line BC matches with the slope of y=-3.5x-15.

the slope of the line DE matches with the slope of y=-0.5x-3.

the slope of the line HI matches with the slope of y=1.25x+4.

the slope of the line LM matches with the slope of y=5x+9.

and the slopes of the lines FG and JK do not match with any of the functions given.

Thus,

$B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15$

$D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3$

$H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4$

$L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9$

4 0

3 years ago

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ohaa [14]

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3 years ago

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9 more than 4 times the square of a number.

musickatia [10]

Answer:

4n+9

hope this helps

have a good day :)

Step-by-step explanation:

6 0

3 years ago

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Can anyone slove this question pls

skad [1K]

we have: -10/4 = -2.5 ⇒ -2 > -2.5 > -3

ANSWER : -2 and -3

ok done. Thank to me :>

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2 years ago

Help Me Plz!!!!!!!!!!!!!

Strike441 [17]

Answer:

$a = 6\sqrt 3,\:\:b =12,\:\:c=6\sqrt 2$

Step-by-step explanation:

$by \: 30 \degree - 60 \degree - 90 \degree \: theorem \\ \\ 6 = \frac{1}{2} b\\ \\ 6 \times 2 = b \\ \\ 12 = b \\ \\ \huge\red {\boxed {b= 12}} \\ \\ a = \frac{ \sqrt{3} }{2} b \\ \\ a = \frac{ \sqrt{3} }{2} \times 12 \\ \\ \huge\purple {\boxed {a = 6 \sqrt{3}}} \\ \\ c = \frac{1}{ \sqrt{2} } b \\ \\ c = \frac{1}{ \sqrt{2} } \times 12 \\ \\ c = \frac{ \sqrt{2} }{2} \times 12 \\ \\ \huge\orange {\boxed {c = 6 \sqrt{2}}} \\ \\$

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3 years ago