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givi [52]
3 years ago
10

Can someone help me with this?

Mathematics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

A=10 cm

B=5 cm

Area of square = 25 cm

Step-by-step explanation:

Since that is an equilateral triangle based off the fact that all the sides are the same variable (a) then that means 30 divided by 3 which equals 10. So now we know that a = 10. Lets input that information for the rectangle. Since there are 2 a's that equals 20 and the perimeter is 30 so b has to be 5 since 5 times 2 is 10 plus 20 equals 30. so now we know that a=10 and b=5 so our squares sides are all 5. The area of the square is 25

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1 2.52 Test (CS): Functions
Mumz [18]

Answer:

V=lwh=5 x 9 x 3=135cm3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The ratio of money in Obi's wallet to Rudy's wallet one day was 5:2.
mamaluj [8]

Answer:

The amount that Obi initially had was £20

Step-by-step explanation:

Let

x ----> amount that Obi initially has

y ----> amount that Rudy has

we know that

\frac{x}{y} =\frac{5}{2}

x=2.5y ----> equation A

x-20=y-8

y=x-20+8

y=x-12 ----> equation B

Solve the system by substitution

substitute equation A in equation B

y=(2.5y)-12

solve for y

2.5y-y=12\\1.5y=12\\y=8

Find the value of x

x=2.5(8)=20

therefore

The amount that Obi initially had was £20

4 0
3 years ago
Plz help, I need help!!!!
Otrada [13]

Answer:

56 multiplied by 9 = 504

Step-by-step explanation:

55.8 rounded is 56, and 9.12 rounded is 9

4 0
3 years ago
How do you solve the inequality to -3|2-4u|+5<-13
alukav5142 [94]

Answer:

u=-1

Step-by-step explanation:

-3|2-4u|+5<-13

-6+12u+5<-13

12u<-13-5+6

12u<-12

12u/12<-12/12

u=-1

7 0
3 years ago
List any restrictions on the domain for the equation <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%2B9%7D%7Bx%2B2%7D%20" id="
frez [133]
For a fraction, if the denominator turns to 0, the fraction becomes undefined, and therefore, that's a restriction on a rational.

now, what values of "x" makes the denominator 0?  let's check,

x+2 = 0

x = -2

so, if "x" ever becomes -2, then you'd get

\bf \cfrac{x+9}{x+2}\implies \cfrac{x+9}{-2+2}\implies \implies \cfrac{-2+9}{-2+2}\implies \stackrel{und efined}{\cfrac{7}{0}}

so, the domain, or values "x" can take on safely, are any real numbers EXCEPT -2.
7 0
3 years ago
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