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lisabon 2012 [21]

3 years ago

11

Can you please answer without sending links because I can’t open them, thanks!

1 answer:

3241004551 [841]3 years ago

8 0

Answer:

B. C. E.

Explanation:

In answer B, 4b means four of that variable, so you would add them all together, same goes with answer C. In answer E, you would use the "multiply by the reciprocal" rule, which basically means to keep the b as it is, change division to multiplication, and flip $\frac{1}{4}$ .

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Alicia needs to purchase carpet to cover a triangular space. The space has a base length of 10 ft and a height of 6.4 ft.

Vika [28.1K]

Answer:

Alicia needs 32ft² of carpet to cover the space

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2 years ago

If there was 17 boys and21 girls what would be the ratio

Lady bird [3.3K]

17 boys to 21 girls
17 boys/21 girls
17 boys:21 girls

8 0

3 years ago

The half-life of uranium-238 is 4.5x10^9 years. The half-life of uranium-234 is 2.5x10^5 years. How many times greater is the ha

lara31 [8.8K]

<u>Answer:</u>

Half-life of uranium-238 $18 \times 10^{3} \text { times greater }$ that of uranium-234

<u>Explanation:</u>

Half time of uranium-238 = $4.5\times 10^9$ years

Half time of Uranium-234 = $2.5\times 10^5$ years

To find how much times greater the half life of uranium-238 is from uranium-234

= $\frac{\text { Half life of Uranium-238 }}{\text { Half time of Uranium - 234 }}$

= $\frac{4.5 \times 10^{9}}{2.5 \times 10^{5}}$

= $18 \times 10^{3} \text { times greater }$

Hence Uranium-238 is $18 \times 10^{3} \text { times greater }$ than Uranium-234

4 0

4 years ago

Divide 224 - 918 + al2 by a6.

Korvikt [17]

Answer:

It will be -6 I hope that helps

Step-by-step explanation:

3 0

3 years ago

Peter the optician sees 16 patients one Tuesday. 4 are neither short nor long sighted. The total number of short sighted patient

Juliette [100K]

Answer:

No of patients having longsightedness = 5

No of patient having shortsightedness = 7

Patients requiring varifocal lenses will be =12

Step-by-step explanation:

Total no of patients = 16

As

No of patients having neither long nor shortsightedness = 4

Let

No of patients having longsightedness = x

So

No of patient having shortsightedness = $x+2$

Equation becomes

$x+(x+2)+4= 16\\2x+6 = 16\\2x= 16-6\\2x= 10\\x= \frac{10}{2} \\x= 5$

Now

No of patients having longsightedness = x= 5

No of patient having shortsightedness = $x+2$ = $5+2$

= 7

Patients requiring varifocal lenses will be = 5+7

= 12

3 0

3 years ago