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2 years ago

Find it!?!?!??!!!!!

Mathematics

2 answers:

Dvinal [7]2 years ago

5 0

Answer:

(0,-4)

Step-by-step explanation:

Whatever point the line is crossing across the y-axis is the y-intercept

VashaNatasha [74]2 years ago

3 0

Answer:

-4

Step-by-step explanation:

It intercepts the y axis at -4

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What is the length of the altitude of the equilateral triangle below?

balandron [24]

$\bf \textit{height or altitude of an equilateral triangle}\\\\ h=\cfrac{s\sqrt{3}}{2}~~ \begin{cases} s=\stackrel{length~of}{a~side}\\ \cline{1-1} s=8\sqrt{3} \end{cases}\implies h=\cfrac{8\sqrt{3}\cdot \sqrt{3}}{2}\implies h=\cfrac{8\sqrt{3^2}}{2} \\\\\\ h=4\cdot 3\implies h=12$

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2 years ago

Can someone answer this? Thanks a lot

Komok [63]

Answer:

Step-by-step explanation:

Use math-way go quick and easy answers without step by step explanations, and use math-soup-calculator for step by step explanation

3 0

2 years ago

Read 2 more answers

The temperature of coffee served at a restaurant is normally distributed with an average temperature of 160 degrees Fahrenheit a

posledela

Answer:

The 20th percentile of coffee temperature is 155.4532 degrees Fahrenheit.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 160 degrees

Standard Deviation, σ = 5.4 degrees

We are given that the distribution of temperature of coffee is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.2

$P( X < x) = P( z < \displaystyle\frac{x - 160}{5.4})=0.2$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 160}{5.4} = -0.842\\\\x = 155.4532$

Thus,

$P_{20}=155.4532$

The 20th percentile of coffee temperature is 155.4532 degrees Fahrenheit.

7 0

2 years ago

Find the interval(s) of upward concavity on this accumulation function.

maxonik [38]

$\displaystylef(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\=2x\tan(x)+\ln(\cos^2(x))$

Hope this helps.

6 0

2 years ago

Is 100 in greater than 3yd 1ft

wolverine [178]

1 yard = 36 inches

3 * 36 = 108

1 foot = 12 inches

108 + 12 = 120

So 3 yards and 1 foot is equal to 120 inches.

120 is greater than 100 so 3 yards and 1 foot is greater than 100 inches.

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2 years ago

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Find it!?!?!??!!!!!​

Find it!?!?!??!!!!!