Which alegbraic expression models the given word phrase?

Mathematics

1 answer:

bixtya [17]3 years ago

3 0

40 fewer than a number which is T would be

C 40-t

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A triangle has vertices at (3, 2), (-4,4), and (-2,5). What are the coordinates of the vertices of the image

MrMuchimi

Answer:

The triangle is translated 2 units down and 4 units to the right.

Step-by-step explanation:

7 0

3 years ago

Table of values 3x+2y=24

Mekhanik [1.2K]

3
x
+
2
y
>
24
3
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+
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y
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Solve for
y
y
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y
>
−
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>
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+
12
Use the slope-intercept form to find the slope and y-intercept.
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3
2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
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2
+
12
-
3
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2
+
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.
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4 0

3 years ago

Add. <br> (−2+6x^3−3x^2)+(4x^3−5+x^2)

Gemiola [76]

Answer:

10x^3 - 2x^2 -7

Step-by-step explanation:

To subtract polynomials, you want to subtract the biggest exponent from the first polynomial to the second

so 6x^3 + 4x^3 and that would be 10x^3

then -3x^2 + x^2 and the negative is larger so it would be -2x^2

and finally -2 + -5 and that is -7

hope this helps! :)

7 0

2 years ago

Each model represents one whole. In which model does the shaded section represent the product of 5/8 and 3/8?

Mnenie [13.5K]

We can’t answer this question without a photo of the model. Can you post a photo of the question?

8 0

4 years ago

Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD

Ugo [173]

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = $\sqrt{75+25}$ = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that $\frac{1}{2} * BF = \frac{1}{2} * 5 * \sqrt{75}$ and BF= $0.5\sqrt{75}$ . Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is $A= BF * \frac{AB+CD}{2} = 0.5 \sqrt{75} * \frac{5+10}{2} = 18.75 \sqrt{3} cm^{2}$

3 0

3 years ago