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Mathematics

2 answers:

Stella [2.4K]3 years ago

8 0

I think the answer would be 3quarts,2 cups+? = 12 quarts, 3 cups (C)

Katarina [22]3 years ago

7 0

12quarts 3cups-3quarts 2cups
And
3quarts 2cups+?=12quarts 3cups

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What constant term should be used to complete the square? x 2 - 5x + _____ = 7

amm1812

The quadratic equation in its generic form is:
ax2 + bx + c
To complete squares we must add the following term:
(b / 2) ^ 2
The equation is:
ax2 + bx + c + (b / 2) ^ 2
We have the following equation:
x ^ 2 - 5x + k = 7
By completing squares we have:
x ^ 2 - 5x + (-5/2) ^ 2 = 7 + (-5/2) ^ 2
Rewriting:
x ^ 2 - 5x + 6.25 = 7 + 6.25
Answer:
A constant term should be used to complete the square is:
6.25

4 0

3 years ago

Antoinette solves the linear equation 3(x-3)+2x+9 = 2X+ 2(x-1) using the steps shown below.

taurus [48]

Answer:

Antoinette's error

Option A) She distributed incorrectly.

Step-by-step explanation:

We are given the following information in the question:

Antoinette solves the linear equation

$3(x-3)+2x+9 = 2x+ 2(x-1)$

Antoinette made mistake in step 1 that is she did not distributed correctly.

If she would have distributed correctly, then she would have got the following solution:

$3(x-3)+2x+9 = 2x+ 2(x-1)\\\text{Step 1: }3x-9+2x+9 = 2x + 2x-2\\\text{Step 2: }5x = 4x -1\\\text{Step 3: }5x-4x = -1\\\text{Step 4: }x = -1$

Substituting x = -1 in the given equation, we gwt:

$3(-1-3)+2(-1)+9 = 2(-1)+ 2(-1-1)\\-12-2+9 = -2-4\\-6=-6\\\text{which is true}$

Antoinette's error

Option A) She distributed incorrectly.

5 0

3 years ago

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26•26^8 i’m not sure how to solve this

alexira [117]

Answer:

26^9

Step-by-step explanation:

7 0

3 years ago

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What is the distance rounded to the nearest tenth between the points(-5, 3) and (1, -1)?

kakasveta [241]

7.2 units is the distance between the two points (-5,3) and (1, -1).

The formula used for determining the distance between two points is shown below. In case this equation is used for finding the distance between the origin and a point, the value of x1 and y1 would just be equal to zero.

distance = $\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$