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3 years ago

What is the length of z? (Round to the nearest tenth)

Mathematics

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

Step-by-step explanation:

$9^{2} + 6^{2} = c^{2}$

81 + 36 = $c^{2}$

117 = $c^{2}$

c = $\sqrt{117}$

c = 10.8

c = z

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Gina's literacy bucket weighs 6 pounds. Her novel weighs 1 4/6 pounds, and her chrome book weighs 2 1/6 how much would her bucke

Paraphin [41]

Answer:

$2\frac{1}{6}$ pounds.

Step-by-step explanation:

We have been given that Gina's literacy bucket weighs 6 pounds. Her novel weighs 1 4/6 pounds, and her chrome book weighs 2 1/6.

To find weight of bucket after taking out the two items, we will subtract weight of each item from 6 pounds as:

$\text{Weight of bucket}=6-1\frac{4}{6}-2\frac{1}{6}$

Let us convert mixed fractions into improper fractions as:

$1\frac{4}{6}=\frac{6\cdot 1+4}{6}=\frac{10}{6}\\\\2\frac{1}{6}=\frac{6\cdot 2+1}{6}=\frac{12+1}{6}=\frac{13}{6}$

$\text{Weight of bucket}=6-\frac{10}{6}-\frac{13}{6}$

$\text{Weight of bucket}=\frac{6\cdot 6}{6}-\frac{10}{6}-\frac{13}{6}$

$\text{Weight of bucket}=\frac{36}{6}-\frac{10}{6}-\frac{13}{6}$

Combine numerators:

$\text{Weight of bucket}=\frac{36-10-13}{6}$

$\text{Weight of bucket}=\frac{13}{6}$

$\text{Weight of bucket}=2\frac{1}{6}$

Therefore, the weight of the bucket is $2\frac{1}{6}$ pounds.

6 0

3 years ago

Issac drove 100 miles at a speed of 50 mph and then 120 miles at a speed of 30 mph what was his average speed for the entire tri

Lina20 [59]

Answer:

The average speed is 36.67 m/h

Step-by-step explanation:

100/50 = 2

120/30 = 4

Total time = 6

Total distance = 220 miles

220/6 = 36.67

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3 years ago

Which one of the following numbers will appear farthest to the left on a number line?

gulaghasi [49]

I think not sure I think -2 but if I had the whole answer to your question I might know more

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3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

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3 years ago

What is the square root of 2

Aliun [14]

Answer: is 1.41421 hope this helps :)

6 0

3 years ago

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