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aleksandr82 [10.1K]

3 years ago

10

Nathan cut a rectangular tile in half for his kitchen floor design. The tile was NOT a square. He made once cut along a diagonal

from one vertex to another vertex. Classify the TWO triangles resulting from the cut by their angles and their side angles.

2 answers:

zhenek [66]3 years ago

6 0

They will both be right, scalene triangles.

Harlamova29_29 [7]3 years ago

5 0

Answer with Explanation:

It is Given that Nathan cut a rectangular tile in half for his kitchen floor design.

As,the cut is made along a diagonal from one vertex to another vertex.

If you cut rectangle along through any of two diagonals,

you will get two triangles which are congruent.

→∵ Opposite sides are Congruent.

→Diagonal is common.

So,two triangles will be congruent by SAS.

Triangles categorized as

1. Scalene triangle

2. Right Triangle

The Two Triangles named in a single way as Scalene Right Triangle.

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Select the correct answer.

Irina18 [472]

D. About 122 units I just took the test

4 0

2 years ago

A number of squares are connected in a row to form a rectangle. The table shows the relationship between the number of squares,

svp [43]

Answer:

✅The table can be represented by the equation, $y = 4x + 4$

✅The relation is a function.

✅The graph of the function is linear.

Step-by-step explanation:

✍️The first statement: The table can be represented by the equation, $y = 4x + 4$ .

To check if the first statement is correct let's find the equation that can represent the table by finding the slope (m) and y-intercept (b).

Using two pairs, (1, 8) and (2, 12),

Slope = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 8}{2 - 1} = \frac{4}{1} = 4$ .

Substitute, x = 1, y = 8, and m = 4 into y = mx + b to solve for b.

Thus:

8 = (4)(1) + b

8 = 4 + b

Subtract 4 from each side

8 - 4 = b

4 = b

Plug in the values of b and m into y = mx + b.

Thus, the equation that represents the table would be:

$y = 4x + 4$ .

✅Therefore, the first statement is correct.

✍️The second statement: The graph of the function is non-linear.

This is NOT TRUE because the equation of the function, $y = 4x + 4$ , represents the equation of a linear graph in the slope-intercept form. When graphed, it will give us a straight line.

✍️The Third statement: The relation is a function.

This is TRUE because each input value (x-value) has exactly one output value (y-value).

✍️The Fourth statement: The rate of change is NOT constant.

This standby is NOT TRUE.

The rate of change is the slope (m) that we have calculated above to be 4. Between any two pairs, the rate of change remains 4.

Therefore, this statement is not correct.

✍️The Fifth statement: The graph of the function is linear.

This is TRUE. As stated earlier, from the equation generated, it is safe to say that the equation represents graph of a linear function. The graph will be a straight line graph.

5 0

3 years ago

Which one is correct?

larisa86 [58]

Answer:

c is the answer

Step-by-step explanation:

4 0

3 years ago

Which of the following describes the function x^3 − 8?

klemol [59]

Answer:

(A)

Step-by-step explanation:

A)The degree of the function is odd, so the ends of the graph continue in opposite directions. Because the leading coefficient is positive, the left side of the graph continues down the coordinate plane and the right side continues upward.

6 0

3 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

3 years ago