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2 years ago

7

What is the lowest common multiple of 24, 8 and 16.

1 answer:

tester [92]2 years ago

4 0

Answer:

48 is the answer

Step-by-step explanation:

I hope this helps because I can't show the fomula because this is a PC

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Emelia earns $8.00 per hour plus a gas allowance of $3.50 per day at her job. How much does Emilia's job pay in a day when she w

Bogdan [553]

Answer: Required expression = 8h +3.50

Emilia's job pay $ 43.5 in a day when she works 5 hours.

Step-by-step explanation:

Given: Hourly earning = $8.00

gas allowance per day = $3.50

Pay per day = (Hourly earning ) x (Number of hours ) + (gas allowance per day)

Let h = number of hours

Pay per day = 8h +3.50

If Emily worked for 5 hours, then put h= 5, we get

Pay per day = 8(5) +3.50

= $ 40 +3.50

= $ 43.5

Required expression = 8h +3.50

Emilia's job pay $ 43.5 in a day when she works 5 hours.

7 0

2 years ago

-5(x + 3) - 4x = 6(2x + 3) - 12

MaRussiya [10]

Answer:x=-1

Step-by-step explanation:

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2 years ago

Find the coordinates of the vertices of the figure after the given transformation: T<0,7>

zaharov [31]

Answer: B

Step-by-step explanation:

Why did historians choose to study this topic?

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2 years ago

Mr Denmark has 28 students in his class he wants to see them in seven equal rows how many students will be nitro

finlep [7]

4 people would be in each row of there were 7 rows

4 0

2 years ago

solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything

vova2212 [387]

a.

$XA^{-1}=A^3$

$(XA^{-1})A=A^3A$

$X(A^{-1}A)=A^4$

$X=A^4$

b.

$AXB=(BA)^2$

$A^{-1}(AXB)B^{-1}=A^{-1}(BA)^2B^{-1}$

$(A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}$

$X=A^{-1}(BA)^2B^{-1}$

c.

$(A^{-1}X)^{-1}=(AB^{-1})^{-1}(AB^2)$

$X^{-1}A=(BA^{-1})(AB^2)$

$X^{-1}A=B(A^{-1}A)B^2$

$X^{-1}A=B^3$

$(XX^{-1})A=XB^3$

$XB^3=A$

$X(B^3(B^3)^{-1})=A(B^3)^{-1}$

$X=A(B^3)^{-1}$

d. Not totally sure what the equation is supposed to be, but I guess it's

$ABXA^{-1}B^{-1}=A$

$ABX(BA)^{-1}=A$

$((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)$

$X=(AB)^{-1}A(BA)$

$X=(B^{-1}A^{-1})A(BA)$

$X=B^{-1}(A^{-1}A)(BA)$

$X=(B^{-1}B)A$

$X=A$

7 0

2 years ago