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elena55 [62]
3 years ago
12

Can the side lengths of 12, 15, and 13 form a triangle? Yes No

Mathematics
2 answers:
grandymaker [24]3 years ago
8 0
My answer is yes! The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. So the sum of 2 sides, must be greater than the third... I hope this helps please rate and thanks
tino4ka555 [31]3 years ago
4 0
Yes. 12 and 13 would be the legs, and 15 would be the hypotenuse. This would form a right triangle.
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Jose purchased 4/9 pound of peanut and 7/11 pounds of raisins find the total weight of his purchase
azamat

Answer:

The total weight of his purchase is 1.08 pounds

Step-by-step explanation:

To find the total weight of his purchase, we sum the weight of each of his purchases.

He purchased:

4/9 pound of peanut.

7/11 pounds of raisins

Total:

The least common multiple between 9 and 11 is 99.

Then

\frac{4}{9} + \frac{7}{11} = \frac{11*4 + 9*7}{99} = \frac{107}{99} = 1.08

The total weight of his purchase is 1.08 pounds

8 0
3 years ago
Which of the following numbers can be expressed as repeating decimals? 3 over 7, 2 over 5, 3 over 4, 2 over 9
Maurinko [17]

Hello!

Let's think about the denominators we have. A seventh goes on forever in no specified order. A fifth an a fourth have set decimal values, 0.2 and 0.25, so they are not repeating decimals. A ninth is 0.111111..., it repeats like this forever in a specified order. Therefore, 2/9, or 0.222222...., would be a repeating decimal.

Therefore, our answer is \boxed {\frac{2}{9}}.

I hope this helps!

6 0
3 years ago
Find the quadratic function y=​f(x) whose graph has a vertex ​(−3​,4​) and passes through the point ​(−7​,0). Write the function
olganol [36]

Answer:

Step-by-step explanation:

This is a parabola since a quadratic is a parabola.  The standard form for a parabola is y = ax² + bx + c

but before we do that, we will use the vertex form, since it will make our work easier at the beginning.  

First and foremost, when we plot the vertex and the given point, the vertex is higher up than is the point; that means that this parabola opens upside down, and its vertex form will be

y=-|a|(x - h)² + k

The absolute value is out in front of the a, so we know that the value of a is positive, but the quadratic itself is negative (upside down) and we will find that math takes care of that negative that needs to be out front.  So we need to solve for a by filling in the x, y, h, and k values from the point and the vertex:  x = -7, y = 0, h = -3, k = 4

0 = a(-7 - (-3))² + 4 and

0 = a(-7 + 3)² + 4 and

0 = a(-4)² + 4 and

0 = a(16) + 4 and

0 = 16a + 4 and

-4 = 16a so

a=-\frac{1}{4}

Now that we know a, we can plug it back into the vertex form and then put it into standard form from there.

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

Now we will FOIL out what's inside the parenthesis to get

y=-\frac{1}{4}(x^2+6x+9)+4

Simplify by distributing the -1/4 into the parenthesis:

y=-\frac{1}{4}x^2-\frac{3}{2}x-\frac{9}{4}+4

Combine like terms to get

y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{7}{4}

And there you go!

5 0
3 years ago
How do you do this problem
Sliva [168]

Answer:


Step-by-step explanation:

sec= x

cross multiply

60 mb x 2

120 mb/29 sec

6 0
3 years ago
What are the coordinates of this and radius!
LiRa [457]

-1,1 I already told u, nut

3 0
3 years ago
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