50 POINT PLS HELP

How do you use numerical expressions to solve real-world problems?

PilotLPTM [1.2K]

There are many ways to do this, if you wanted to write a numerical expression for: Lauren had $7 she babysits for three hours and earned $8.50 per hour, you would write 7+3x8.50=M. It is much shorter and makes much more sense to write this expression than to write a whole story

5 0

3 years ago

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Please help me please thank you so much in advance

Phoenix [80]

Answer:i think its 90 degrees ?

Step-by-step explanation:

5 0

3 years ago

4 – (2y + 5) = 3(1 – 4y) show steps please

Lelu [443]

Answer/Step-by-step explanation:

Let's solve your equation step-by-step.

4−(2y+5)=3(1−4y)

Step 1: Simplify both sides of the equation.

4−(2y+5)=3(1−4y)

4+−1(2y+5)=3(1−4y)(Distribute the Negative Sign)

4+−1(2y)+(−1)(5)=3(1−4y)

4+−2y+−5=3(1−4y)

4+−2y+−5=(3)(1)+(3)(−4y)(Distribute)

4+−2y+−5=3+−12y

(−2y)+(4+−5)=−12y+3(Combine Like Terms)

−2y+−1=−12y+3

−2y−1=−12y+3

Step 2: Add 12y to both sides.

−2y−1+12y=−12y+3+12y

10y−1=3

Step 3: Add 1 to both sides.

10y−1+1=3+1

10y=4

Step 4: Divide both sides by 10.

10y

10

=

4

10

y=

2

5

Answer:

y= 2/5

yw

6 0

3 years ago

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A man rides a bicycle a distance of 48 miles in a certain amount of time. If the man increased his speed by 2 miles per hour, he

seropon [69]

Recall your d = rt, distance = rate * time.

so, in his usual speed, say is hmm "r" mph, so in his usual "r" speed, he rolls on for the 48 miles and it takes him "t" hours to get there.

now, if he increases his speed by 2 mph, then his new speed is "r + 2", and he arrives there 4 hours earlier, so if he took "t" hours going at "r" speed, then when he's going faster at "r + 2", he only takes "t - 4" hours, for the same 48 miles.

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{usual speed}&48&r&t\\
\textit{faster speed}&48&r+2&t-4
\end{array}
\\\\\\
\begin{cases}
48=rt\implies \frac{48}{r}=\boxed{t}\\
48=(r+2)(t-4)\\
----------\\
48=(r+2)\left( \boxed{\frac{48}{r}}-4 \right)
\end{cases}
\\\\\\
48=(r+2)\left( \cfrac{48-4r}{r} \right)\implies 48r=(r+2)(48-4r)
$

$\bf 48r=48r-4r^2+96-8r\implies 0=-4r^2+96-8r
\\\\\\
4r^2+8r-96=0\implies r^2+2r-24=0
\\\\\\
(r+6)(r-4)=0\implies r=
\begin{cases}
-6\\
\boxed{4}
\end{cases}$

it cannot be a negative value, since is just a forward speed rate, so can't be -6.

8 0

3 years ago

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A boy pulls the handle of a lawn mower at an angle of 30° to the horizontal with a force

ella [17]

If the mower is pulled over a flat horizontal surface, then first get the component of the force in the direction of the mower's displacement:

(120 N) cos(30°) = 60 √3 N ≈ 103.923 N

Then the work done by the boy on the mower over a distance of 12.8 m is

(103.923 N) (12.8 m) ≈ 1330.2 Nm ≈ 1330 J

7 0

3 years ago