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3 years ago

6

construction workers resurface parts of the N1 Highway.They complete 2km per day.How many kilometres are completed in total afte

r the first 3 days

1 answer:

Alex73 [517]3 years ago

6 0

6 kilometers I think

I'm not sure

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Please help me asap. I need an answer and full explanation. 1st person to complete that gets Brainliest. Question: Amelia likes

Zielflug [23.3K]

Answer:

22 cards

Step-by-step explanation:

Set up a ratio.

$\frac{34}{29+x} =\frac{2}{3}$

$58+2x=102$

$2x=44$

$x=22$

Check: Add 22 to 39. 22+29=51

34/51 = 2/3

Have a great day

3 0

3 years ago

A grapefruit is 8% heavier than an orange, and an apple is 10% lighter than the orange.

likoan [24]

Weight of an grapefruit=weight of an orange+8% weight of an orange
weight of an apple=weight of an orange -10% weight of an orange

a.<span>By what percentage is the grapefruit heavier than the apple?
We should find the connection between grapefruit and an apple. We know the connection between the weight of a grapefruit and an orange, we know the connection between an orange and an apple, so this means we know the connection between a grapefruit and an apple.
</span>
weight of an grapefruit=weight of an <span>orange+8% weight of an orange
</span>weight of an orange=weight of an apple<span> +10% weight of an apple
</span>
-> weight of an grapefruit=weight of an apple+10% weight of an apple + 8%(weight of an apple+10% weight of an apple)= weight of an apple + 18% weight of an apple + 2% weight of an apple= <span>weight of an apple + 20% weight of an apple

</span><span>b.By what percentage is the apple lighter than the grapefruit?
</span>weight of an grapefruit=weight of an apple + 20% weight of an apple<span>
</span>
-> The apple ts 20% lighter than the grapefruit.

8 0

3 years ago

Read 2 more answers

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$

7 0

3 years ago

Please help it’s a pic

Evgesh-ka [11]

Answer:

58°

Step-by-step explanation:

<MQR = <XQL (vertically opposite angles)

28 + 5b = 70 - 2b

28 - 70 = -2b - 5b

-42 = -7b

-42/-7 = b

6 = b

<MQR = 28 + 5b

= 28 + 5(6)

= 28 + 30

= 58°

7 0

3 years ago

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GEOMETRY HELP PLEASE!!!!!!!!!!!

Setler [38]

I have the same problem. if i find the answer i will give it to you

8 0

3 years ago