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AnnZ [28]
3 years ago
9

ricardo has 2 cases of video games with the same number of games in each case. he gives 4 games to his brother. ricardo has 10 g

ames left. how many video games were in each case?
Mathematics
2 answers:
Margaret [11]3 years ago
5 0
10+4=14, 14/2=7, 7 would be the answer
insens350 [35]3 years ago
3 0
First, you have to add 10+4=14, then you have to divide 14 by 2 to get 7
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a local school raised 5687.50 for the terry fox foundation. Hua raised 0.8% of this total. how much money did Hua raise?
zlopas [31]
You multiply the $5687.50 by 0.8 to get the total of $4550.
So Hua raised $4550 out of the $5687.50.
6 0
3 years ago
The sum of a number and 10​
adell [148]

Answer:

10 + x

Step-by-step explanation:

Sum means add

Hope this helps you :)

pls give me brainlest ;)

3 0
3 years ago
I’ll mark brainlest! pls help! :/
alexandr1967 [171]

Answer:

9\sqrt{2}

Step-by-step explanation:

d = \sqrt{(5-(-4)^2+(5-(-4)^2 \\

d = \sqrt{(9)^2+(9)^2

d = 9\sqrt{2}

6 0
3 years ago
Read 2 more answers
The base of an isosceles right triangle is 30 cm. What is its area?
dlinn [17]

Answer:

A = 450 cm²

Step-by-step explanation:

Since the right triangle is isosceles then the 2 logs are congruent , both 30 cm and at right angles to each other.

The area (A ) of the triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the perpendicular height )

Here b = h = 30 , then

A = \frac{1}{2} × 30 × 30 = \frac{1}{2} × 900 = 450 cm²

7 0
2 years ago
Write the standard form equation of the ellipse shown in the graph, and identify the foci.
vlada-n [284]

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3)  and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6  and b = 4

The standard equation  becomes

\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1

\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1

Foci  are (h,k+c)  and (h,k-c)

c=\sqrt{a^2-b^2}

Plug in the a=6  and b=4

c=\sqrt{6^2-4^2}

 c=\sqrt{20}

  c=2\sqrt{5}, we know h=-2  and k=-3

Foci  are   (-2,-3+2\sqrt{5})  and  (-2,-3-2\sqrt{5})

Option A is correct

6 0
3 years ago
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