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1 ) During his morning commute to work in rush hour traffic, Justin's average speed was 30 mi/h. During his afternoon commute ba

Hitman42 [59]

Answer:45 mi/h

Step-by-step explanation:

Recall: Speed = distance / time

Let the time taken for the morning journey be $t_{1}$ and the time taken for the afternoon journey be $t_{1}$ ,and the distance covered be d

that means for the first journey,

30 = $\frac{d}{t_{1}}$

d = 30 $t_{1}$

Also for the afternoon journey,

d = 60 $t_{2}$

Equating the two , since the same distance is being covered , we have

30 $t_{1}$ = 60 $t_{2}$

that is

$t_{1}$ = $\frac{60t_{2}}{30}$

$t_{1}$ = 2 $t_{2}$

Also ,

total distance covered = 30 $t_{1}$ + 60 $t_{2}$

Average speed = total distance / total time

= 30 $t_{1}$ + 60 $t_{2}$ / $t_{1}$ + $t_{2}$

Recall that $t_{1}$ = 2 $t_{2}$ , substitute this into the formula for average speed , then we have

Average speed = 30(2 $t_{2}$ ) +60 $t_{2}$ / 3 $t_{2}$

Average speed = 120 $t_{2}$ / 3 $t_{2}$

Therefore :

Average speed = 40 mi/hr

4 0

3 years ago

CHECK MY ANSWERS PLEASE!!

lukranit [14]

Answer:

1. A) OPS is congruent to RSU

2. C) RSP is congruent to TSU

Step-by-step explanation:

1. Since OQ and RT are parallel lines that are cut across by the transversal line, NU, therefore, angle OPS and angle RSU are corresponding angles formed. Corresponding angles are congruent.

✅Therefore: A) OPS is congruent to RSU. OPTION A IS TRUE.

❌OPS and RSP are consecutive interior angles. They are not congruent. They are supplementary. So, option B is NOT TRUE.

❌OPS and OPN are linear pair angles. Linear pair angles are not congruent. They are supplementary. Therefore, OPTION B IS NOT TRUE.

OPS and TSU are also not congruent.

2. Angle RSP and angle RSU are vertical angles. Vertical angles are congruent. Therefore, the statement that IS TRUE is:

C) RSP is congruent to TSU

5 0

3 years ago

How do you complete problem number 4?

brilliants [131]

(I'm going to translate y' to dy/dx as it makes it easier to read for me, you could change it back if you wanted.)
$x \frac{dy}{dx} -y=0$
$\frac{dy}{dx} = \frac{y}{x}$
$\int \frac{1}{y} dy= \int \frac{1}{x} dx$ (separate the variables)
$\ln y=\ln x +c$
$\ln y = \ln kx$ (by letting c = ln k and using log laws)
$y=kx$ (raise everything to power e)
$3=k\times1 \implies k=3$ (applying boundary conditions)
Particular solution: $y=3x$

3 0

4 years ago

Write the absolute value inequality in the form

yawa3891 [41]

Answer:

$\large\boxed{|x+7|\geq2}$

Step-by-step explanation:

Look at the picture.

$|x-a|\geq b\\\\a=\dfrac{-9+(-5)}{2}=\dfrac{-14}{2}=-7\\\\b=-9-(-7)=-9+7=-2\\b=-5-(-7)=-5+7=2\\\\|x-(-7)|\geq2\\\\|x+7|\geq2\\\\Check:\\\\|x+7|\geq2\iff x+7\geq2\ \vee\ x+7\leq-2\qquad\text{subtract 7 from both sides}\\\\x+7-7\geq2-7\ \vee\ x+7-7\leq-2-7\\\\x\geq-5\ \vee\ x\leq-9\qquad CORRECT\ :)$

4 0

4 years ago

In what order must you preform the operations indicated by a negative rational exponent?

Svet_ta [14]

The operations referred to are likely

• raising to a power (the numerator of the exponent)

• taking a root (the denominator of the exponent)

• finding a reciprocal (because the exponent is negative)

They can be performed in any convenient order. It often works well to deal with small positive integers, so if one or more of these operations lets you proceed with a small positive integer for the remaining operations, that would be the one you'd perform first.

_____

A computer performing operations with a negative rational exponent may do so using logarithms. That is, the log of the base will be multiplied by the exponent, then the antilog found. The exponent itself will likely be treated as a floating point number, unless coding specifically indicates otherwise.

4 0

3 years ago