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AleksandrR [38]
2 years ago
13

Find the common difference of the arithmetic sequence 4, 10, 16,

Mathematics
2 answers:
DIA [1.3K]2 years ago
7 0

Answer:

10it is 10

Step-by-step explanation:

julia-pushkina [17]2 years ago
5 0
I could be wrong but i think it’s 10, hope it helps
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Can someone plz help plz show work
aliya0001 [1]

Answer:

<h2>LA = 24 ft²</h2>

Step-by-step explanation:

In the base we have the right triangle. Use the Pythagorean theorem for to calculate the hypotenuse:

h^2=4^2+3^2\\\\h^2=16+9\\\\h^=25\to h=\sqrt{25}\\\\h=5\ ft

The Lateral Area is the sum of the areas of three rectangles:

4\ ft\ \times\ 2\ ft\to A_1=(4)(2)=8\ ft^2\\\\3\ ft\ \times\ 2\ ft\to A_2=(3)(2)=6\ ft^2\\\\5\ ft\ \times\ 2\ \ft\to A_3=(5)(2)=10\ ft^2

The Lateral Area:

LA=A_1+A_2+A_3\to LA=8+6+10=24\ ft^2

8 0
3 years ago
Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to
aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x 10^{8} square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x 10^{4} Km.

Surface area of the star = 4n^{2}

Where n is the radius of the star.

So that;

n = \frac{ 1.8083 * 10^{4} }{2}

   = 0.90415 x 10^{4}

n =  0.90415 x 10^{4} Km

Thus,

Surface area = 4 x (0.90415*10^{4} )^{2}

                     = 326994889

Surface area = 3.2700 x 10^{8} km^{2}

Therefore, the surface area of the star is 3.2700 x 10^{8} square kilometres.

4 0
3 years ago
Lizzie rolls two dice. What is the probability that the sum of the dice is:
zhenek [66]

Answer:

A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A)  = \dfrac{12}{36} = \dfrac{1}{3}

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)}  Count 1

Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

P(B)  = \dfrac{15}{36} = \dfrac{5}{12}

C. Divisible by 3 and less than 7:

P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

P(A \cap B) = \dfrac{7}{\text{36}}

5 0
2 years ago
Please help with this
solmaris [256]
I think you should do this equation
4 0
3 years ago
Read 2 more answers
HELPP. I DON’T GET HOW TO DO THIS!!!! You have to write 2 equations and solve using method of substitution or method of eliminat
ipn [44]

Answer:

Can you make the picture bigger if you do i can solve it for u my co puter donset zoom in so..

Step-by-step explanation:

6 0
2 years ago
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