Please help solve this SAT math problem

Answer: Look at picture

Answer:

a.) 0,51

b.) 51%

c.) 1/100

d.) 1%

e.) 3/20

f.) 0,15

Step-by-step explanation:

[tex]a.) \: \frac{51}{100} = 51 \div 100 = 0.51[/tex]

[tex]b.) \: \frac{51}{100} = 51\%[/tex]

[tex]c.) \: 0.01 = \frac{1}{100} [/tex]

[tex]d. )\: 0.01 = \frac{1}{100} = 1\%[/tex]

[tex]e.) \: 15\% = \frac{15}{100} = \frac{3}{20} [/tex]

[tex]f.) \: 15\% = \frac{15}{100} = 0.15[/tex]

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Step-by-step explanation:

There are 12 months in a year so,

114/12= 9.5$/month

t=9.5m

Since T represents the total cost, we want to express it in terms of months by multiplying m which refers to the number of months by the monthly fee which i have already calculated.

Using this equation we can find out the answer to your previous question.

t=9.5m

but m=3×12+5=41 months (3 years 12 months each)

t=9.5(41)

t=389.5 $

Answer:

32 Bags.

Step-by-step explanation:

Answer:

The answer is d.

Step-by-step explanation:

[tex]x^{2} -7x+12=0\\(x-3)(x-4)=0\\x-3=0; x=3\\x-4=0; x=4\\(3,4)[/tex]

Answer:

fog =gofthen it will be equal

Answer:

See below

Step-by-step explanation:

Definitions are:

Use the old exp-log trick and properties of the logarithm to rewrite the limit as

[tex]\displaystyle \lim_{n\to\infty} \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)^{\frac xn}[/tex]

[tex]\displaystyle = \exp\left[\lim_{n\to\infty} \frac xn \ln \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac nk\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\left(\frac nk\right)^2\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac1{k/n}\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\frac1{\left(k/n\right)^2}\right)\right][/tex]

The first limit converges to 0, since n! asymptotically behaves like nⁿ, so ln(nⁿ/n!) → ln(1) = 0.

The two remaining sums converge to definite integrals:

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x + \frac1{\frac kn}\right) = \int_0^1 \ln\left(x + \frac1y\right) \, dy = \frac{(x+1) \ln(x+1)}x[/tex]

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x^2 + \left(\frac1{\frac kn}\right)^2\right) = \int_0^1 \ln\left(x^2 + \frac1{y^2}\right) \, dy = \frac{2\tan^{-1}(\sqrt x)}{\sqrt x} + \ln(1+x)[/tex]

It follows that

[tex]f(x) = \exp\left[x\left(0 + \dfrac{(x+1)\ln(x+1)}x - \dfrac{2\tan^{-1}(\sqrt x)}{\sqrt x} - \ln(x+1)\right)\right][/tex]

[tex]f(x) = \exp\left[\ln(x+1) - 2\sqrt x \tan^{-1}(\sqrt x)\right][/tex]

[tex]f(x) = (x+1) e^{-2\sqrt x \tan^{-1}(\sqrt x)}[/tex]

By computing f'(x) and f''(x), it's easy to show that f'(x) ≤ 0 and f''(x) ≥ 0 for all x > 0. So f(x) is decreasing and f'(x) is increasing, and

• (A) f(1/2) ≥ f(1) is true

• (B) f(1/3) ≤ f(2/3) is false

• (C) f'(2) ≤ 0 is true

• (D) f'(3)/f(3) ≥ f'(2)/f(2)   ⟺   f'(3)/f'(2) ≥ f(3)/f(2)   ⟺   (something larger than 1) ≥ (something smaller than 1) is true

Answer:

nope, it’s actually 176,000. the 901 is very high if u compare it to 499 which is the lowest number to round down to.

Step-by-step explanation:

Gradient = Slope.

Use two points on the line to find the slope.

Slope = change in y over change in x

Points: (-4,0) and (0,-4)

Slope = (-4 - 0) / (0 - -4)

Slope = -4/4 = -1

Slope = -1

Answer:

B. x - 10

Step-by-step explanation:

x² - 13x + 30 = (x - 3)(x - 10)

x² + (a + b)x + ab = (x + a)(x + b)

Answer:

4x^3

Step-by-step explanation:

GCF is the Greatest Common Factor

12x^6 + 4x^4 + 28x^3

= 4x^3(3x^3 + x + 7)

GCF is 4x^3

Answer:

5

Step-by-step explanation:

6 / 2 = 15 / l

3 = 15 / l

l = 15 / 3

l = 5

Answer: 5/9 is the simplified fraction for 25/45 by using the GCD or HCF method. Thus, 5/9 is the simplified fraction for 25/45 by using the prime factorization method.

Answer:

[tex]x=20[/tex]

Step-by-step explanation:

[tex]x^2=8(8+42)[/tex]

[tex]x^2=8*50[/tex]

[tex]x=20[/tex]

~

Answer:

Let's solve your equation step-by-step.

p=(0.5)(2.718282)−5

Step 1: Simplify both sides of the equation.

p=(0.5)(2.718282)−5

p=1.359141+−5

p=(1.359141+−5)(Combine Like Terms)

p=−3.640859

p=−3.640859

Answer:

p=−3.640859

Step-by-step explanation:

Answer:divide both sides of the equation by 10.4

Step-by-step explanation:

Answer:

its d

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

i took the test

The height of the prism is 49.2 feet

The given parameters are:

The area of the trapezoid is:

Area = 0.5 * (Sum of opposite sides) * Height

So, we have:

1280 = 0.5 * 52 * Height

This gives

1280 = 26 * Height

Divide both sides by 26

Height = 49.2

Hence, the height of the prism is 49.2 feet

Read more about trapezoids at:

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Answer:

x=5

Step-by-step explanation:

The door of the ramp must begin at a distance of 25.72ft in order to raise the ramp by 5ft.

Data;

To solve this problem, this is a right angle triangle and we can use trigonometric ratio SOHCAHTOA here.

Since we have the value of angle and opposite and solving only for adjacent, we can use tangent of the angle.

[tex]tan \theta = \frac{opposite}{adjacent} \\[/tex]

Let's substitute the values into the formula above and solve

[tex]tan 11 = \frac{5}{x} \\x = \frac{5}{tan 11} \\x = 25.72ft[/tex]

The door of the ramp must begin at a distance of 25.72ft in order to raise the ramp by 5ft.

Learn more on trigonometric ratio here;

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Answer: B 0.4

Step-by-step explanation:

50-20=30

0.4 is 40%

40% of 50 is 20

Answer:

Angles 7 and 8

Step-by-step explanation:

They are next to each other, and share a common side.

Answer:

I guess how many times out of x they will do something.

for example how many times out of 10 will bob ask out his crush vs Joe

Step-by-step explanation:

There isn't really an easy way to measure this.

The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

The scale of measurement deals with how the variables are defined and categorized.

There are 4 types of scales of measurement;

1. Nominal scale

2. Ordinal scale

3. Interval scale

4. Ratio scale

We know that the Ordinale scale is used for variables in ranked order but the difference between them could not be determined.

Hence, The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

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9514 1404 393

Answer:

  19

Step-by-step explanation:

The prime factorization of 4221462 is ...

  4221462 = 2·3·7·100511

so, the 1-digit divisors are ...

  1, 2, 3, 6, 7

The total of these numbers is 19.

Answer:

4,000

600

160

12

Step-by-step explanation:

When using partial products, you basically break up the number into ones, tens, hundreds, thousands, etc.

So the partial products of 1,643 x 4 will be,

1,000 x 4

600 x 4

40 x 4

3 x 4

The partial products of the multiplication problems are,

4000

600

160

12

The answer will be 4,772.

The partial products of the multiplication problems are,

4000, 2400, 160 and 12

Multiplication is a method of finding the product of two or more numbers

When using partial products, you basically break the number into ones, tens, hundreds, thousands, etc.

So the partial products of 1,643 x 4 will be,

1,643 can be written as 1000+600×40+3.

Now multiply with 4.

(1000+600×40+3)×4

1,000 x 4+600 x 4+40 x 4+3 x 4

4000+2400+160+12

Hence the partial products of the multiplication problems are,

4000, 2400, 160 and 12 and The answer will be 4,772.

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Answer:

Sergio found 4 pennies on the ground and her sister said that she found twice as many as him

When his sister said that she found twice as many, sergio added 2 to his number of pennies instead of multiplying by 2

That's where Sergio made a mistake and got the answer wrong