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VikaD [51]
3 years ago
5

5. Andre is solving the equation 4(x + 3) = 7. He says, “I can subtract from each side to get

Mathematics
1 answer:
Vinil7 [7]3 years ago
8 0
4(x+3)=7
The mistake is with his final answer, he wrote down 4x=1.

When you do 4(x)4(3), the answer is completely different from what Andre said. He did wrong calculations.

4x3=12

7-12= -5

Which means it should be 4x=-5 as the final answer.
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4. For graduation, Diamond was able to buy her first car for $24,000. Her parents gave her a present and made a
s344n2d4d5 [400]

Answer:

$403.15

Step-by-step explanation:

Principal loan amount is the total amount minus down-payment:

Principal=24000-1000\\\\=23000

Knowing that n=12\times 6=72,P=23000, r=7.99\%/12=0.006658, the monthly payments can be calculated using the formula:

M=P[\frac{r(1+r)^n}{(1+r)^n-1}]\\\\=23000\frac{(0.006658(1.006658)^{72}}{1.006658^{72}-1}\\\\=403.15

Hence, the monthly payment is $403.15

3 0
3 years ago
What is negative six one third equal negative nine three fourths
11Alexandr11 [23.1K]

Answer:

Do you have a picture of the problem so I can see it

8 0
3 years ago
HELP ME QUICK: Suppose AB has one endpoint at A(0, 0). If (5, 3) is the midpoint of AB, what are the coordinates of point B?
DiKsa [7]

Answer:

<h3>           B(10, 6)</h3>

Step-by-step explanation:

If P is midpoint of AB and:  A(0,\,0)\,,\quad P(5,\,3)\,,\quad B(x_B,\,y_B)

then:

x_P-x_A=x_B - x_P\qquad\quad\ \wedge\qquad y_P-y_A=y_B - y_P\\\\ 5-0=x_B-5\qquad\quad\wedge\qquad 3-0=y_B -3\\\\ x_B=5+5\qquad\qquad\wedge\qquad\ \ y_B=3+3 \\\\ {}\quad x_B=10\qquad\qquad\wedge\qquad\qquad \ y_B=6

5 0
3 years ago
DOES ANYONE KNOW THE ANSWERS TO THESE PROBLEM, PLZ HELP!
Virty [35]

Answer:

Q.1.

Sum of angles of a triangle = 180°

∠A + ∠B + ∠c = 180

(3x - 9) + (4x + 15) + 90 = 180

3x - 9 + 4x + 15 + 90 = 180

7x + 96 = 180

7x = 84

x = 12

∠A = 3x - 9 = 3(12) - 9 = 36 - 9 = 27°

∠B = 4x + 15 = 4(12) + 15 = 48 + 15 = 63°

∠A = 27°

∠B = 63°

Q.2.

using \ pythagoras \ theorem \\x^2 = 7^2 + (7\sqrt{3} )^{2} = 49  + 49(3) = 196\\x = \sqrt{196}  = 14

sin y = \frac{opposite}{hypotenuse} = \frac{7\sqrt{3} }{14}   = \frac{\sqrt{3} }{2} \\\\y = sin^{-1} (\frac{\sqrt{3} }{2} ) = 60

x = 14

y = 60°

8 0
3 years ago
A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C
harina [27]

Answer:

The answer is (B) but with 2 as the last item in the matrix, instead of 1.

Step-by-step explanation:

The grand topic is MATRICES.

In Matrices, there are rules to follow when creating a matrix - which is a 'row by column' arrangement of objects or figures.

Rule 1: Identify your rows and columns.

What is a row? A row is a 'left to right' arrangement of items; also called a horizontal arrangement of items. What is a column? A column is an 'up to down' arrangement of items; also called a vertical arrangement of items.

Rule 2: Start your arrangement by rows.

Before reaching the instruction in the question, I already arranged the data for cables on the rows and the data for wire colors on the columns. You should follow this rule always; from the first information given in the question.

Rule 3: Nomenclature

This is the naming of a given matrix. In naming a matrix, consider the rows first before the columns. For example, a matrix with three rows and four columns with be called a 3 by 4 matrix and not a 4 by 3 matrix. In the case of this question, there is an equal number of rows and columns, so even if you counted the columns first, it wouldn't be noticed. Nevertheless, always count the rows first.

Having arranged the figures in a 3 by 3 matrix,

The full data on Cable A appears on the first row. The full data on Cable B appears on the second row. The full data on Cable C appears on the third row.

The full data on Black wires appears on the first column. The full data on White wires appears on the second column. The full data on Red wires appears on the third column.

So the answer is \left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right]

Kudos!

5 0
3 years ago
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