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Sphinxa [80]
3 years ago
11

Josh earned $80 to use for back-to-school shopping. So far he has spent $24 on a backpack and $32 on new shoes. Think abt the yo

u would need to figure out how much money Josh has left
Mathematics
1 answer:
Lady_Fox [76]3 years ago
7 0

Answer:

24$

Step-by-step explanation:

$80 - $32 - $24 = $24

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Blake and 3 friends meet for lunch . his​
____ [38]

Answer:

if this is finish the problem then

"his friends decided to order pizza for lunch. one of his friends said they wanted to eat 5 slices. the other one said he wanted 3 and the last one said they must wanted 7 while little Blake just wanted 2. how many boxes of pizza would Blake have to order."

MY answer: zero because Blake made them pay for their own boxes. they didn't want to at first but then they saw the knife Blake was holding in his hand and decide that if they want to live they must buy their own pizza and give him some or theirs

7 0
3 years ago
Read 2 more answers
5 + 14 x y = 9 x y - 5
Anon25 [30]

Answer:

xy = -2

Step-by-step explanation:

To solve:

5 + 14xy = 9xy - 5

=> First subtract 9xy from each side of "="

5 + 5xy = -5

=> Now, subtract 5 from each side:

5xy = -10

=> Finally divide 5 on each side:

xy = -10/5

xy = -2

Hope this helps!

4 0
3 years ago
Read 2 more answers
Area of a triangle with points at (-9,5), (6,10), and (2,-10)
Ann [662]
First we are going to draw the triangle using the given coordinates. 
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Distance from point A to point B:
d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}
d_{AB}= \sqrt{(6+9)^2+(10-5)^2}
d_{AB}= \sqrt{(15)^2+(5)^2}
d_{AB}= \sqrt{225+25}
d_{AB}= \sqrt{250}
d_{AB}=15.81

Distance from point A to point C:
d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}
d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}
d_{AC}= \sqrt{11^2+(-15)^2}
d_{AC}= \sqrt{121+225}
d_{AC}= \sqrt{346}
d_{AC}= 18.60

Distance from point B from point C
d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}
d_{BC}= \sqrt{(-4)^2+(-20)^2}
d_{BC}= \sqrt{16+400}
d_{BC}= \sqrt{416}
d_{BC}=20.40

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
s= \frac{AB+AC+BC}{2}
s= \frac{15.81+18.60+20.40}{2}
s= \frac{54.81}{2}
s=27.41

Finally, to find the area of our triangle, we are going to use Heron's formula:
A= \sqrt{s(s-AB)(s-AC)(s-BC)}
A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}
A= \sqrt{27.41(11.6)(8.81)(7.01)}
A=140.13

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0
3 years ago
The following are the weights, in pounds, of some dogs at a kennel: 36, 45, 29, 39, 51, 49
MAXImum [283]
What is the question...??
5 0
3 years ago
write the equation of a line that is perpendicular to the given line and that passes through the given point. y-4=5/2(x+3) ; (-7
faltersainse [42]
The gradient of the perpendicular line would be the negative reciprocal of the original line. Therefore the gradient of the perpendicular line would be -2/5x.
Since we know y=mx+c, ∴y=-2/5x+c. Sub in the x and y values of the given point and we get that c=26/5.

The perpendicular equation would be y=-2/5x+26/5.
I hope I got this right.
7 0
4 years ago
Read 2 more answers
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