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3 years ago

Does anyone know this?

Mathematics

1 answer:

weeeeeb [17]3 years ago

7 0

Answer:

BC = 10.24

Step-by-step explanation:

You can use the Pythagoras theorem for this question

$a^{2}$ + $b^{2}$ = $c^{2}$

a = 8cm

b = ???

c = 13cm

We will solve for B (length BC)

$a^{2}$ + $b^{2}$ = $c^{2}$

$8^{2}$ + $b^{2}$ = $13^{2}$

Rearrange for b

$b^{2}$ = $13^{2}$ - $8^{2}$

$b^{2}$ = 105

b = $\sqrt{105}$

b = 10.24

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Answer number 2 for me plz and thank you

anyanavicka [17]

The answer to the question is 2:3

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3 years ago

Read 2 more answers

one light flashes every 2 minutes and another flashes every 7 minutes. If both lights flash together at 1pm, what is the first t

vovangra [49]

Light A flashes every 2 minutes while light B flashes every 7. The goal is to figure out when both lights flash at the same time and then figure out the other part later. You have to find the common multiples of 7 and 2. Meaning that you have to find out which number can they both be multiplied into. In this case, it is 14 because 7 can be multiplied by 2 to get 14 and 2 can be multiplied by 7. So if every 14 minutes they flash together at the same time, now you need to find out what time AFTER 3 will they both flash. You know that they both flashed at 1:00 so they will flash again at 1:14 and 1:28 and so on, but you need to find out when is the soonest they will flash after 3. 2 hours after the 1:00 flash will be 3:00 so that is 120 minutes. 120 minutes divided by 14 minutes comes to 8.5 but you dont need the decimal. Just simply multiply 14 by 8 and that comes to 112 minutes and if you add that to 1:00 you get 2:52. so they flashed at 2:52 so the next time they flash will be 14 minutes after 2:52. Sorry for the super long answer

6 0

3 years ago

7.3 homework help me

olga_2 [115]

Answer:

1. Yes

∆RST ~ ∆WSX

by SAS

2. Yes

∆ABC ~ ∆PQR

by SSS

3. Yes

∆STU ~ ∆JPM

by SAS

4. Yes

∆DJK ~ ∆PZR

by SAS

5. Yes

∆RTU ~ ∆STL

by SAS

5. Yes

∆JKL ~ ∆XYW

by SAS

6. No

7. Yes

∆BEF ~ ∆NML

by SAS

8. Yes

∆GHI ~ ∆QRS

by SSS

9. x=22

10. x=12

Step-by-step explanation:

1. RS/WS=ST/SX and m<RST=m<WSX

2. AB/PQ=8/6=4/3

BC/QR=AC/PR=12/9=4/3

AB/PQ=BC/QR=AC/PR

3. ST/JP=10/15=2/3

SU/JM=14/21=2/3

ST/JP=2/3=SU/JM

and m<TSU=70°=m<PJM

4. DK/PR=8/4=2

JK/ZR=18/9=2

DK/PR=2=JK/ZR

and m<DKJ=65°=m<PRZ

5. RT/ST=UT/LT

and m<RTU=m<STL

6. KL/YW=20/18=10/9

JL/XW=36/24=3/2

KL/YW=10/9≠3/2=JL/XW

7. BF/NL=24/16=3/2

BE/NM=39/26=3/2

BF/NL=3/2=BE/NM

and m<EBF=m<MNL

8. GH/QR=32/20=8/5

HI/RS=40/25=8/5

GI/QS=24/15=8/5

GH/QR=HI/RS=GI/QS=8/5

9. x/33=18/27

Simplifying the fraction on the right side of the equation:

x/33=2/3

Solving for x: Multiplying both sides of the equation by 33:

33(x/33)=33(2/3)

x=11(2)

x=22

10. x/16=9/12

Simplifying the fraction on the right side of the equation:

x/16=3/4

Solving for x: Multiplying both sides of the equation by 16:

16(x/16)=16(3/4)

x=4(3)

x=12

4 0

3 years ago

3x²-30x+9=0<br> PLEASE HELP<br> COMPLETING THE SQUARE WILL MARK BRAINLIEST

nlexa [21]

Answer:

i have done the cutting of x's exponent

Step-by-step explanation:

= 3x - 30+9 =0

= 3x - 21 =0

= 3x= 21

= x = 21/3

x= 7 this the answer

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3 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$