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3 years ago

Find the missing side lengths. Leave your answers as radicals in simplest form

Mathematics

1 answer:

konstantin123 [22]3 years ago

4 0

X=y=29 divided by square root of 2.
Hope this helped!!!
Sorry I could not give you the exact value.
I guess it is 20.506

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Write the decimal as a fraction 0.12

Juli2301 [7.4K]

The answer is 12/100. This is because the 12 is a 100 decimal place.

7 0

3 years ago

Read 2 more answers

Which exponential equation is equivalent to the logarithmic equation below? c = ln 2

Sophie [7]

$\text{A logarithmic lemma: } \\ \text{It follows that: } \log_a a = 1$
$\text{Following this property, we can see that: } a^{\log_{a} a} = a$
$\text{Furthermore, we can create more general properties: } a^{\log_a b} = b$

$c = \log_e 2 \Rightarrow e^{c} = e^{\log_e 2} \\ \text{Using our third property, we see that: } e^{c} = 2$
$\text{And finally, this works for any } \log_{a} b \left\{a > 0, b \in \mathbb{R}\right\}$

6 0

3 years ago

Samantha has $24 but she wants a pair of Jordan's that are $112. She does chores and she gets paid $5 for each chore. If she doe

natita [175]

Answer:

Check step-by-step.

Step-by-step explanation:

Okay, so. She gets 5 dollars and she does 15 chores. You would multiply it.

5 * 15 = $75

Then you would subtract 112 and 75.

112 - 75 = 37.

She will get $75. Then, she will need to gather $37

6 0

2 years ago

Adding/Subtracting rational expressions

Leona [35]

7]
6/(x-1)-5x/4
subtracting the above we put the fraction under the same denominator:
6/(x-1)-5x/4
multiplying the denominators we get:
4(x-1)
thus subtracting we get:
6/(x-1)-5x/4
=(4*6-5x(x-1))/[4(x-1)]
=[24-5x^2+5x]/(4x-4)
Answer:
(-5x^2+5x+24)/(4x-4)

9]
3/(x+7)+4/(x-8)
the common denominator is:
(x+7)*(x-8)=(x+7)(x-8)
thus adding the fractions we put them under the same denominator as follows:
[3(x-8)+4(x+7)]/[(x+7)(x-8)]
=[3x-24+4x+28]/[(x+7)(x-8)]
=(7x+4)/[(x+7)(x-8)]

5 0

3 years ago

All the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxe

AveGali [126]

Answer:

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

Step-by-step explanation:

Given that all the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over.

This implies that boxes are in multiples of 12

i.e. 12, or 24, or 36 or....

Let us say this as 12m for m an integer

Now additional boxes arrived =60

Total number of boxes now = $12m+60$

This when arranged in stacks of 14,nothing left out

Or this is multiple of 14

$12m+60 = 14n$ for some positive integer n.

By trial and error we find that if 60 is to be divided by 14, 24 should be added and 24 is a multiple of 12.

So we say originally 24 boxes were there now 84 boxes there.

Hence both options are right.

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

3 0

3 years ago