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2 years ago

A dog walks at a speed of 3mi/hr for 2 hours. How far (distance) did the dog walk? *

Mathematics

1 answer:

Nikolay [14]2 years ago

6 0

Answer:

6 miles

Step-by-step explanation:

After one hour the dog will have walked three miles after two hours the dog will have walked three more miles for a total of 6 miles.

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Find the distance between points (5,1) and (3,4) to the nearest tenth

Charra [1.4K]

Answer:

3.6 units (nearest tenth)

Step-by-step explanation:

distance formula

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

Thus, distance between the 2 points

= √[(5 - 3)² + (1 - 4)²]

= √[ 2²+ ( - 3)²]

= √13

= 3.6 units (nearest tenth)

5 0

2 years ago

X + y = 17 -x + 3y 7-8 :

ladessa [460]

Answer:

it could be x= 10 and y=7? I think bc 17 can be split up into 10 and 7 or 9 and 8 etc.

4 0

2 years ago

Find the value of x. 2x 120° 700 A. 55 B. 35 C. 25 D. 10

monitta

Answer:

for this question you need to take the 12p and 700 and add then dived by 2x which then gives you your answer

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2 years ago

Read 2 more answers

The product of 6g2 and -2g3 is equals to

ryzh [129]

Answer:

We conclude that:

$\left(6g^2\:\right)\times \left(-2g^3\right)=-12g^5$

Step-by-step explanation:

Given

6g²

-2g³

The product of 6g² and -2g³ can be determined by multiplying each other such as:

$(6g^2\:)\times \:(-2g^3)$

Apply the rule: a(-b) = -ab

$\left(6g^2\:\right)\times \left(-2g^3\right)=-6g^2\times \:\:2g^3$

Apply exponent rule: $a^b\times \:a^c=a^{b+c}$

$=-6g^{2+3}\times \:2$

$=-6g^5\times \:2$

$=-12g^5$

Therefore, we conclude that:

$\left(6g^2\:\right)\times \left(-2g^3\right)=-12g^5$

8 0

2 years ago

The value of y varies directly with x. when y = 7.5, x = 5. what is the value of y when x = 80?.

Ksenya-84 [330]

Answer:

Step-by-step explanation:

120 = y. the ratio is 3:2

80/2 is 40

40 times 3 is 120

120:80 = 3:2

4 0

3 years ago

A dog walks at a speed of 3mi/hr for 2 hours. How far (distance) did the dog walk? *​

A dog walks at a speed of 3mi/hr for 2 hours. How far (distance) did the dog walk? *