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3 years ago

a truck traveled 8 hours at an average speed of 55 mph. how far did the truck travel? (distance = rate x time)

Mathematics

2 answers:

Bezzdna [24]3 years ago

5 0

Answer:

440 Miles.

Step-by-step explanation:

In your question, you have mentioned that d=r•t, and in this scenario, plugging int he values would give us D=8•55.

The distance of the truck is r•t.

Nonamiya [84]3 years ago

3 0

Answer:

8*55=440 miles per hr.

Step-by-step explanation:

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Can someone Help with this question

inn [45]

Answer:

B 12

Step-by-step explanation:

manuel ate 1/3X
His brother ate 1/4X
leftover is 5
total is X

manuel+brother+leftover=X

x/3+x/4+5=x

7x+60/12=x

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5 0

3 years ago

Are you cyclist travels at a rate of 1.8 km/min. how far will the cycle is travel in 48 minutes

Tpy6a [65]

Answer:

86.4 km

Step-by-step explanation:

Since the cyclist travels 1.8km in a minute, he will travel 48 times that in 48 minutes:

$48*1.8 = 86.4$

6 0

3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$