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tensa zangetsu [6.8K]
2 years ago
11

Please help please help

Mathematics
1 answer:
BigorU [14]2 years ago
6 0

Answer:

(1/2, 1&1/2)

Step-by-step explanation:

the x-axis, which is the horizontal axis, is always written first. the y-axis, vertical, is written second. because the points give are spaced out every other line, it shows that there fractional numbers must be used

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Help what is the answer to number 1 plz
Serga [27]
The property shown there is the distributive property.
It states that a(b + c) = ab + ac
Or a(b - c) = ab - ac

8 0
3 years ago
Prove that if n is an odd integer, then n^2 + 1 is an even integer.
MrRa [10]

Answer:

n=3

9+1=10

n=1

1+1=2

n=5

25+1=26

a odd interger squared is a odd interger so if you add one its even

Hope This Helps!!!

4 0
3 years ago
In the equation y = 6.7x, what is the constant of proportionality
devlian [24]

The constant in an equation would be the number multiplied to x,

In the equation y - 6.7x, the constant would be 6.7

4 0
3 years ago
In which quadrants do solutions for the inequality y ≤ 2/7x +1 exist?
viktelen [127]

All four quadrants

<h2>Explanation:</h2>

We have the following inequality:

y \leq \frac{2}{7}x+1

So the first step we need to perform is to plot the line:

y = \frac{2}{7}x+1

If \ x=0 \\ \\ y=\frac{2}{7}(0)+1 \\ \\ y=1 \\ \\ \\ If \ y=0: \\ \\ 0=\frac{2}{7}(x)+1 \\ \\ x=-\frac{7}{2}=-3.5

So the line passes through the points:

(0,1) \ and \ (-3.5,0)

To find the shaded region, let us take a point, namely, the origin and test it in the inequality:

y \leq \frac{2}{7}x+1 \\ \\ 0\leq \frac{2}{7}(0)+1 \\ \\ 0\leq 1 \ True!

Since this is true, then the shaded region includes this point. This is shown below and <em>as you can see the solutions exist in all four quadrants.</em>

<h2>Learn more:</h2>

Inequalities: brainly.com/question/12890742

#LearnWithBrainly

3 0
3 years ago
Read 2 more answers
Which congruency statements could be correct for the figures? Check all that apply. MNOP ≅ STUV MNPO ≅ TSVU NPOM ≅ VUTS OPNM ≅ T
jeka94

Answer:

NPOM ≅ VUTS

OPNM ≅ TUVS

Step-by-step explanation: Given: two quadrilaterals having verticals P, N, O,M and S,T,V,U are congruent,  where, OM is congruent or equal to TS and in quadrilaterals NPOM and  VUTS-

since, the condition and, side UV=side  OM   follow for the above quadrilateral. (According to the figure)

then we can say according to the property of quadrilateral, their corresponding sides must be congruent. so they are congruent.

7 0
3 years ago
Read 2 more answers
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