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alex41 [277]
4 years ago
13

What number : Increased by 130% is 69 ?

Mathematics
1 answer:
e-lub [12.9K]4 years ago
4 0
<h2>Answer:<u> 158.7 is the answer</u></h2><h2><u>or it can be</u> <u>89.7</u></h2>

Step-by-step explanation:

<h2><u>69 + Percentage increase = </u></h2><h2><u>69 + (130% × 69) = </u></h2><h2><u>69 + 130% × 69 = </u></h2><h2><u>(1 + 130%) × 69 = </u></h2><h2><u>(100% + 130%) × 69 = </u></h2><h2><u>230% × 69 = </u></h2><h2><u>230 ÷ 100 × 69 = </u></h2><h2><u>230 × 69 ÷ 100 = </u></h2><h2><u>15,870 ÷ 100 = </u></h2><h2><u>158.7</u></h2><h2><u /></h2><h2><u /></h2><h2><u>69 increased by 130% = 158.7 </u></h2><h2><u>Absolute change (actual difference): </u></h2><h2><u>158.7 - 69 = 89.7</u></h2>
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The diagram below represents the net of a triangular prism. What is the surface area of the triangular prism?
AleksandrR [38]

Answer:

(C)

Step-by-step explanation:

The diagram is divided ninto 5 parts that is rectangles ABCD, CHGD, GFEH and  triangles CIH and DGJ.

Area of rectangle ABCD= l{\times}b

=12{\times}6=72cm^{2}

Area rectangle CDHG=  l{\times}b

=12{\times}8=96cm^{2}

Area rectangle HGFE= l{\times}b

=10{\times}12=120cm^{2}

Area of triangles CIH and DIG= 2{\times}\frac{1}{2}{\times}b{\times}h

=6{\times}8=48cm^{2}

Surface area of the triangular prism=Area of rectangle ABCD+Area rectangle CDHG+Area rectangle HGFE+Area of triangles CIH and DIG

=72+96+120+48=336cm^{2}

Hence, option C is correct.

3 0
3 years ago
Can I please have help?
givi [52]
-12+-12=-24
-6+-6=-12
-11+-11=-22
-10+-10=-20
-9+-9=-18
-8+-8=-16
-7+-7=-14
-5+-5=-10
-4+-4=-8
-3+-3=-6
-1+-1=-2
1+1=2
2+2=4
3+3=6
4+4=8
5+5=10
6+6=12

8 0
3 years ago
Wht is the substitution of x+y=9 and x=2y
Rama09 [41]
I hope this helps you



2y+y=9



3y=9



y=3


x=2y=2.3=6
3 0
3 years ago
I need help with math!!
fomenos

Answer:

ok the anser is 39

Step-by-step explanation:

ypu do bacic equagens

7 0
3 years ago
Help I don’t understand
OverLord2011 [107]
Rotation about a point does not change any dimensions. C'D' = CD = 2.8 units.
8 0
3 years ago
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