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Vladimir79 [104]
3 years ago
10

The Healey family drove 192 miles in 4.5 hours. How many miles could they drive at this rate in 3 hours? A-64 miles

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

192 miles D

------------- = ----------

4.5 hours 3 hours

I did 192 x 3=576

and then:

576/ 4.5=128

Hope this helps!

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Alla [95]

Answer:

50+90+3x+1=180.....................

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3 years ago
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Can somebody help me please
serious [3.7K]

Answer:

-7 and 21

Step-by-step explanation:

y-2y=7 (sub x=7 to the equation)

-y=7

y=-7 (not sure of its correct)

if y=-7,

x-2(-7)=7 (sub y=-7 into equation)

x-14=7

X=7+14

X=21

8 0
3 years ago
The height of a trapezoid is 8in. and its area 96 in. (2nd Power) One base of the trapezoid is 6 inches longer that the other ba
Jlenok [28]
The lengths of the bases would be 3 and 9
Step by Step Description:

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3 years ago
Cards labeled 5, 6, 7, 8, and 9 are in a stack. A card is drawn and not replaced. Then, a second card is drawn at random. Find t
Rom4ik [11]

Answer:

1/10

Step-by-step explanation:

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6 0
3 years ago
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
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