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2 years ago

PLSSSS HELPPPPPPPPP MEEEEEEEEEEEEEE

Mathematics

1 answer:

Ghella [55]2 years ago

6 0

<h3>Answer: 4/25</h3>

=========================================================

Explanation:

Add up the first two frequencies to get 14+18 = 32

The table shows that the person rolled either a "1" or a "2" a total of 32 times, which is the amount of rolls getting less than "3".

This is out of the 200 rolls total.

32/200 = (8*4)/(8*25) = 4/25

In decimal form, this would be 4/25 = 0.16 which is exact.

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F(x)=2x^2-5;find when f(x)=-7

Daniel [21]

Answer:

x=i or no solution exists depending on the user's grade

Step-by-step explanation:

f(x) = 2x^2-5

-7 = 2x^2-5

-2 = 2x^2, x^2=-1, x=i

6 0

3 years ago

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-8z+(4.5)+3.5z+7y-1.5

bogdanovich [222]

Answer:

$\boxed{\sf{7y+3-4.5z}}$

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

$\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}$

First, thing you do is remove parentheses.

$\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}$

Solve.

Then, you combine like terms.

$\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}$

Add/subtract the numbers from left to right.

-8z+3.5z=-4.5z

Rewrite the problem down.

$\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}$

Solve.

4.5-1.5=3

$\Longrightarrow: \boxed{\sf{7y+3-4.5z}}$

Therefore, the correct answer is 7y+3-4.5z.

I hope this helps, let me know if you have any questions.

6 0

2 years ago

Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2

Arturiano [62]

Answer:

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0

3 years ago

the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?

madam [21]

Answer:

About 0.6548 grams will be remaining.

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$

5 0

3 years ago

Solve 3x-2/4 - 2x-5/3 = 1+x/6

elena-14-01-66 [18.8K]

Answer:

x = 3.8

hope it's helpful ❤❤❤

THANK YOU.

5 0

3 years ago

Read 2 more answers