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3 years ago

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