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wlad13 [49]
3 years ago
10

Find QD if QB=21 and PC=11

Mathematics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

The answer to your question is: 21

Step-by-step explanation:

Data

AB = BC = CD

BQ = 21

QD = ?

PC = 11

Process

Here there are 2 triangles

ΔQCB and ΔQCD

These triangles share one side QC

And one side measures the same BC = CD

And the ∠ BCQ = ∠ DCQ

Then these triangles are congruents  SAS

if QB = 21 then QD = 21

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\huge \pink\star\huge{ \pink{\bold {\underline {\underline {\red{Answer }}}}}}

➭Doing it all using mental math

<h2 /><h2><u>Step 1:</u></h2>

1037 + 4 \\  =1041

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3 years ago
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4 years ago
Solve (x – 3)2 = 49. Select the values of x. –46 -4 10 52
Fudgin [204]

Answer: x = 55/2

Step-by-step explanation: (x)(2) (-3)(2) = 49                                                       2x-6+6 = 49+6                                                                                                               2x/2 = 55/2                                                                                                                   x = 55/2

I know that my answer isn't a choice for one of the values of x but that's the result that I got when I solved (x–3)2 = 49.

6 0
3 years ago
Using the graph of f(x) = log2x below, approximate the value of y in the equation 22y = 5.
Naddika [18.5K]
The correct question is
Using the graph of f(x) = log2x below, approximate the value of y in the equation 2^(2y) = 5

we have 
2^(2y) = 5--------------> applying base 2 logarithm both members
2y*log2(2)=log2(5)
log2(2)=1
then
2y=log2(5)
y=[log2(5)]/2

using the graph
for  x=5  the approximate  value of log2(5) is 2.3
see the attached figure
so
y=[log2(5)]/2--------> y=[2.3]/2----------> y=1.15

the answer is
the approximate value of y is 1.15

4 0
3 years ago
Read 2 more answers
The height of a triangle is 12 inches and the base of the triangle is 17 inches. What is the area of the triangle?
beks73 [17]

Answer:

102 inches squared

Step-by-step explanation

A = (b x h) / 2

A = (12 x 17) / 2

= 102

8 0
3 years ago
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