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alekssr [168]
1 year ago
9

Solve for b: 3b – 5 = 43

Mathematics
2 answers:
Thepotemich [5.8K]1 year ago
7 0

we have the following:

3b-5=43

solving for b:

\begin{gathered} 3b-5=43 \\ 3b=43+5 \\ b=\frac{48}{3} \\ b=16 \end{gathered}

therefore, the answer is 16

laila [671]1 year ago
6 0

Answer: b=16

Step-by-step explanation: The file

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What is the value of h(-7)<br><br> h(x) =|3x| -1
Art [367]

Answer:

20

Step-by-step explanation:

h(-7) just means that inside the x value, you replace the x with a -7.

h(-7) = |3x| - 1

h(-7) = | 3(-7)| - 1

h(-7) = | -21 | - 1

NOTE: -21 is inside |x|, those straight lines mean absolute value. No matter what, any negative number would turn into a positive one. And if it’s posititve, then it stays positive.

h(-7) = 21 - 1

h(-7) = 20

6 0
4 years ago
What is the answer? <br>x+y=15 <br>-2x+5y=-2​
zavuch27 [327]
Multiply the first equation by 2

2x + 2y =30

now we’ll add the second equation to the first equation

2x + 2y =30
-2x+5y= -2

7y=28

divide by 7

y=4

now plug y back into the first equation

x + 4 = 15

subtract 4

x=11
6 0
3 years ago
Help please! i don’t understand how to do this!
atroni [7]

Answer:

x=c+b/a

Step-by-step explanation:

ax-b=c

    +b +b

ax=c+b

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a=c+b/a

4 0
3 years ago
Read 2 more answers
Can someone help meee please
hoa [83]

Answer:

Equation → 5y = 3y + 6

Value of ST = 15

Step-by-step explanation:

From the picture attached,

In right triangles ΔVST and ΔVUT,

Acute angles ∠SVT ≅ ∠UVT [Given]

TV ≅ TV [By reflexive property of congruence]

ΔVST ≅ ΔVUT [Hypotenuse angle congruence of right triangles]

Therefore, corresponding parts of the congruent triangles are congruent.

Therefore, ST ≅ TU

5y = 3y + 6

5y - 3y = 6

2y = 6

y = 3

Therefore, ST = 5y

ST = 5(3)

     = 15

8 0
3 years ago
Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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