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2 years ago

It is now 3:15 pm. Is it possible to drive 135 miles and arrive before 5:00 pm if ypu drove 55 mph? Explain your answer

2 answers:

6 0

Well, how long will it take you to drive 135 miles at 55mph?

at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.

$\bf \cfrac{135}{55}\implies \cfrac{27}{11}\implies 2\frac{5}{11}~hours\implies \textit{about 2hrs, 27 minutes}$

so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.

so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.

Ilia_Sergeevich [38]2 years ago

4 0

From 3 pm to 5 pm is 2 hours, so you have less than 2 hours if you start at 3:15.

In two hours at 55 mph, you can drive 2×55 miles = 110 miles. It will take more than 2 hours to drive 135 miles at that speed.

Is it possible? NO
You cannot complete in less than 2 hours a trip that takes more than 2 hours to do.

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WILL GIVE BRAINLIEST

mafiozo [28]

Answer:

(6x - 1) • (2x + 9)

Step-by-step explanation:

STEP

Equation at the end of step 1

((22•3x2) + 52x) - 9

STEP

Trying to factor by splitting the middle term

2.1 Factoring 12x2+52x-9

The first term is, 12x2 its coefficient is 12 .

The middle term is, +52x its coefficient is 52 .

The last term, "the constant", is -9

Step-1 : Multiply the coefficient of the first term by the constant 12 • -9 = -108

Step-2 : Find two factors of -108 whose sum equals the coefficient of the middle term, which is 52 .

-108 + 1 = -107

-54 + 2 = -52

-36 + 3 = -33

-27 + 4 = -23

-18 + 6 = -12

-12 + 9 = -3

-9 + 12 = 3

-6 + 18 = 12

-4 + 27 = 23

-3 + 36 = 33

-2 + 54 = 52 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and 54

12x2 - 2x + 54x - 9

Step-4 : Add up the first 2 terms, pulling out like factors :

2x • (6x-1)

Add up the last 2 terms, pulling out common factors :

9 • (6x-1)

Step-5 : Add up the four terms of step 4 :

(2x+9) • (6x-1)

Which is the desired factorization

HOPE IT HELPS! :))

8 0

2 years ago

Which system of equation is represented

olga nikolaevna [1]

Answer:

Step-by-step explanation:

First, let us figure out what is graphed and then manipulate each answer to try to match it.

The one with a positive slope is

$y\leq x-2$

The one with the negative slope is

$y\geq 3-2x$

Now that we have our graphed system of inequalities, we can try to match them to an answer.

A. while the x-2 is the same, the other does not have an x, so it cannot match.

B. The second inequality has the wrong sign, so it is impossible for this to be our answer.

C. When each of these are solved for y, you get $y\leq 2x-3\\\\y\leq 2-x$ . Neither of these match so this cannot be correct.

D. When the first inequality is solved you get $y\geq 3-2x$

When the second inequality is solved you get $y\leq x-2$

As both of these answers match our graphed inequalities, D is the correct answer.

4 0

3 years ago

If m∠EBF = 6x + 4 and m∠CBF = 7x – 2, find m∠EBF.

olganol [36]

Answer:

Step-by-step

I hope it's helped

6 0

2 years ago

Read 2 more answers

According to the ASA theorem, which of the following sets of relationships can be used to show that VRT and STR are congruent?

irina [24]

Answer:

C. ∠SRT≅∠VTR and ∠STR≅∠VRT

Step-by-step explanation:

Given:

Quadrilateral is a parallelogram.

RS║VT; RT is an transversal line;

Hence By alternate interior angle property;

∠SRT≅∠VTR

∠STR≅∠VRT

Now in Δ VRT and Δ STR

∠SRT≅∠VTR (from above)

segment RT= Segment RT (common Segment for both triangles)

∠STR≅∠VRT (from above)

Now by ASA theorem;

Δ VRT ≅ Δ STR

Hence the answer is C. ∠SRT≅∠VTR and ∠STR≅∠VRT

7 0

2 years ago

Solve the system of linear equations using multiplication.

Anna71 [15]

Answer:

$(8,-1)$

Step-by-step explanation:

The given system is:

$3x+3y=21$

$6x+12y=36$

Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.

This gives me the system:

$x+y=7$

$x+2y=6$

We could solve the first equation for $x$ and replace the second $x$ with that.

Let's do that.

$x+y=7$

Subtract $y$ on both sides:

$x=7-y$

So we are replacing the second $x$ in the second equation with $(7-y)$ which gives us:

$(7-y)+2y=6$

$7-y+2y=6$

$7+y=6$

$y=6-7$

$y=-1$

Now recall the first equation we arranged so that $x$ was the subject. I'm referring to $x=7-y$ .

We can now find $x$ given that $y=-1$ using the equation $x=7-y$ .

Let's do that.

$x=7-y$ with $y=-1$ :

$x=7-(-1)$

$x=7+1$

$x=8$

So the solution is (8,-1).

We can check this point by plugging it into both equations.

If both equations render true for that point, then we have verify the solution.

Let's try it.

$3x+3y=21$ with $(x,y)=(8,-1)$ :

$3(8)+3(-1)=21$

$24+(-3)=21$

$21=21$ is a true equation so the "solution" looks promising still.

$6x+12y=36$ with $(x,y)=(8,-1)$ :

$6(8)+12(-1)=36$

$48+(-12)=36$

$36=36$ is also true so the solution has been verified since both equations render true for that point.

5 0

2 years ago

Read 2 more answers