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Elis [28]
4 years ago
10

Multiple Response: Please select all correct answers and click "submit."

Mathematics
1 answer:
Marrrta [24]4 years ago
3 0
C.29 and E.1
All other answers when added to 30 equal 60 or more making the statement false.
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Which expression is the best estimate of the product of 7/8 and 1/10
Lera25 [3.4K]
Multiply across 7/80
6 0
3 years ago
What is the name of the rat that is opposite NP?
nignag [31]
<h3>Answer: D) Ray NM</h3>

Ray NP starts at point N as the fixed endpoint and goes off forever in the direction of point P.

The opposite would be to turn 180 degrees and go the other way. So we start at N and instead go off forever toward point M. In either case, the starting point is still the same (point N).

When naming rays, the order is important. Ray NM is different from ray MN. This is because the first letter is the fixed end point (that doesn't go off forever in one direction).

6 0
3 years ago
What is the length of the diagonal if a 10 cm by 15cm rectangle ?
mojhsa [17]
In other words, if a rectangle is 10 cm by 15 cm, what is the length of the diagonal?  Use the Pythagorean Theorem.  a^2+b^2=c^2, where c is the measure of the hypotenuse.
4 0
4 years ago
Read 2 more answers
Find three consecutive positive integers such that the sum of their squares is 2354. What is the largest integer?
Ilya [14]

Answer:

The largest integer is 29.

Step-by-step explanation:

Let the consecutive positive integers are x, x+1 and x+2.

The square of sum of squares of three consecutive positive integers is 2354 such that,

x^2+(x+1)^2+(x+2)^2=2354\\\\x^2+x^2+2x+1+x^2+4+4x=2354\\\\3x^2+6x+5=2354\\\\3x^2+6x- 2349=0

It is a quadratic equation whose solution is given by :

x = 27 and x = -29

First positive integer = 27

Second positive integer = 27+1 = 28

Third positive integer = 27+2 = 29

Hence, the largest integer is 29.

4 0
3 years ago
Po is trying to solve the following equation by completing the square: 49x^2+56x-64 = 0. He successfully rewrites the above equa
ohaa [14]

Answer:

The answer is 91

Step-by-step explanation:

We have the following equality:

49x^2+56x-64=a^2x^2+2abx+b^2-c

Then a must satisfy that 49=a^2. So,

a=7 or a=-7.

1) If a=7, then, it follows for the first equality that 56=14b. Then b=4. Finally, substituting b=4 in the first equality we obtain that -64=16-c. So, c= 80. We conclude that

a+b+c=7+4+80=91.

As the problem states, we only consider values of a greater than zero. Then 91 is the only solution for a+b+c.

5 0
3 years ago
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