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Nina [5.8K]
3 years ago
10

Look at these equations:

Mathematics
1 answer:
OLga [1]3 years ago
6 0

Answer:

A = C + 16

Step-by-step explanation:

Given

A = B + 12 and B = C + 4, then substitute B = C+ 4 into A, that is

A = C + 4 + 12

A = C + 16

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Simplify: Log3 27 = x can anybody help me with this?? Thank you!!
Allushta [10]

Answer:3

Step-by-step explanation:

3 ^x= 27

X=3

5 0
3 years ago
3-4z = -5+8z Can someone help?
noname [10]
3-4z=-5+8z

Add 5 to both sides

8-4z=8z

Add 4z to both sides

8=12z

Divide both sides by 12

z=12/8

Simplify

z=3/2
3 0
3 years ago
Read 2 more answers
Please hellppp ......​
katovenus [111]

Answer:

<h2>BC = 11.9</h2>

Step-by-step explanation:

To solve for BC we use sine

sin ∅ = opposite / hypotenuse

From the question

AC is the hypotenuse

BC is the opposite

So we have

sin 58 = BC / AC

sin 58 = BC / 14

BC = 14 sin 58

BC = 11.87

BC = 11.9 to one decimal place

Hope this helps you

7 0
4 years ago
Read 2 more answers
I begg someone just please helppp
Temka [501]

Answer:

#4.

- 3 < - 2 1/2

If you consider the number line, the numbers increase from left to right. The number on the left is smaller than a number on the right.

- 3 is smaller than -2 1/2, it means -3 is on the left from the -2 1/2 on the number line.

#5.

Jada's score is positive, it means the number reflecting the score is to the right from zero on the number line.

Isabel's score is negative, it means the number reflecting the score is to the left from zero on the number line.

Since Isabel's score is to the left  from the Jada's score, she has lower score than Jada.

With the lowest score, Isabel is the winner.

6 0
3 years ago
This circle is centered at the origin, and the length of its radius is 8. What is
Kitty [74]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x² + y² = 64

____________________________________

\large \tt Solution  \: :

Standard equation of circle is :

\qquad \tt \rightarrow \: (x - h) {}^{2}  + (y - k) {}^{2}  =  {r}^{2}

  • \textsf{h = x - coordinate of centre of circle}

  • \textsf{k = y - coordinate of centre of circle}

  • \textsf{r = radius= 8}

Since the circle is centered at origin, h = k = 0

\qquad \tt \rightarrow \: (x - 0) {}^{2}  + (y - 0) {}^{2}  =  {8}^{2}

\qquad \tt \rightarrow \:  {x}^{2}  +  {y}^{2} = 64

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
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