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3 years ago

Age If you add Natalie's age and Fred's age, the result is 43. If you add Fred's age to 4 times

1 answer:

8 0

Let’s take fred age as x and Natalie’s age as y.
So,
A/Q
X + Y = 43. Let it be eqution 1
and
X + 4Y = 79. Let it be equation 2

Combining equation 1 and 2, we get

( X + Y) + 3Y = 79
43 + 3Y = 79
3Y = 36
Y = 12

Now, evaluating the value of Y in equation 1 we get,

X + 12 = 43
X = 31

Therefore, Fred is 31years old and Natalie is 12 years old.

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Add -3 5/6 + 5 3/7 A.) 1 25/42 B.) 2 8/13 C.)-9 11/42 D.) 1 27/42

Fantom [35]

Answer:

A. 1 25/42

Step-by-step explanation: =-3 5/6 +5 3/7

= (-23/6)+ (38/7)

=(-161+228)/42

=-67/42

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3 years ago

How many solutions does 3(x+5)=-4x+8 have

Mashutka [201]

Answer:

1 solution and that is that x = -1

Step-by-step explanation:

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4 years ago

Solve for x. 4^(×-1)=1

steposvetlana [31]

Answer:

$4^{\left(x-1\right)}=1$

$\Rightarrow\left(x-1\right)\ln \left(4\right)=\ln \left(1\right)$

$\hookrightarrow x=1$

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3 years ago

A swimming pool is a rectangular prism. The pool is 36 feet long and 20 feet wide. What is the total amount of water, in cubic f

Sliva [168]

Answer:

$2880\ ft^{3}$

Step-by-step explanation:

The computation of the total amount of water needed could be determined by using the volume formula which is shown below:

$Volume = L \times W \times D$

where

L = length,

W = width,

and D = depth.

Now putting the values to the above formula

$Volume = (36\ ft) \times (20\ ft) \times (4\ ft) \\\\= 2880\ ft^{3}$

Therefore, the total amount of water needed is $2880\ ft^{3}$

We simply applied the above formula

4 0

3 years ago

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

5 0

1 year ago