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3 years ago

Consider the first three terms of the arithmetic sequence: 7, 15, 23,... Determine d, the common difference.

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

3 0

Answer:

I think 47 and 55

Explanation:

I think that each number is the previous one plus 8;

7+8=15

15+8=23

23+8=31

31+8=39

So, next will be:

39+8=47

47+8=55

Is this what ur looking for?? hope this help:)

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Which is greater 6/12 or 1/3

Inessa [10]

Answer:

1/3 is bigger

Step-by-step explanation:

6/12 might look bigger but its not

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3 years ago

Solve the right triangle. Round decimal answers to the nearest tenth.

AnnZ [28]

Answer:

See solution below

Step-by-step explanation:

According to pythagoras theorem

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3 years ago

What value for Y makes the equation true ? 18y + 4 = 130 {5,6,7,8}

Rufina [12.5K]

Answer:

7 is true

Step-by-step explanation:

if y=5, then 94=130, which is false.

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If y =7, 130=130, which is true

4 0

3 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

PetSmart.com sold the following varieties of dog food in June 2003.

aivan3 [116]

Let x be the number of bags of Nature's Recipe Venison Meal & Rice Canine 20 percent protein, y be the number of bags of Nutro Max Natural Dog Food 27 percent protein and z be the number of bags of PetSmart Premier Oven Baked Lamb 25 percent protein a dog breeder buys. He wants to make 300 pounds of a mix. Then

20x+17.5y+30z=300.

Now

in 20x pounds of Nature's Recipe Venison Meal & Rice Canine 20 percent protein is 0.2·20x=4x pounds of protein;
in 17.5y pounds of Nutro Max Natural Dog Food 27 percent protein is 0.27·17.5y=4.725y pounds of protein;
in 30z pounds of PetSmart Premier Oven Baked Lamb 25 percent protein is 0.25·30z=7.5z pounds of protein;
in 300 pounds of mix containing 22 percent protein is 0.22·300=66 pounds of protein.

Then 4x+4.725y+7.5z=66.

You get a system

$\left\{\begin{array}{l}20x+17.5y+30z=300\\4x+4.725y+7.5z=66.\end{array}\right.$

From the first equation $z=10-\dfrac{2}{3}x-\dfrac{7}{12}y.$ Substitute it into the second equation:

$4x+4.725y+7.5(10-\dfrac{2}{3}x-\dfrac{7}{12}y)=66.$

Simplify it:

$-x+\dfrac{7}{20}y=-9,\\ \\ 20x-7y=180.$

y must be divided by 20, then

y=0, x=9, z=10-6=4;
y=20, x=16, z is not a whole number;
y=40, x=23, z is not a whole number;
y=60, x=30, z=10-20-35<0;
thus, all other possible z will be <0.

Answer: 9 bags with 1st mix, 0 bags with 2nd mix and 4 bags with 3rd mix.

5 0

3 years ago

Consider the first three terms of the arithmetic sequence: 7, 15, 23,... Determine d, the common difference.​

Consider the first three terms of the arithmetic sequence: 7, 15, 23,... Determine d, the common difference.