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Nesterboy [21]
3 years ago
6

What are the first 5 terms of an arithmetic sequence with a common difference of – 0.4 and the first term of 0.7?

Mathematics
2 answers:
vovikov84 [41]3 years ago
5 0
A no is the correct ans .
zimovet [89]3 years ago
5 0
It would be no as a correct answer
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A paperweight is shaped like a triangular pyramid. The base is an equilateral triangle. Find the surface area of the paperweight
stich3 [128]

Answer:

The surface area of the paperweight is SA=8.3\ in^2

Step-by-step explanation:

we know that

The surface area of the triangular prism is equal to the area of its triangular base plus the area of its three triangular lateral faces

so

<em>Find the area of its triangular base</em>

SA=\frac{1}{2}(2)(1.7)

SA=1.7\ in^2

<em>Find the area of its three triangular lateral faces</em>

SA=3[\frac{1}{2}(2)(2.2)]

SA=6.6\ in^2

Adds the areas

SA=1.7+6.6

SA=8.3\ in^2

4 0
4 years ago
Which is greater 0.4 or 0.32
Zanzabum
0.4 is greater than 0.32, because 0.32 is smaller than 0.4.
7 0
3 years ago
Read 2 more answers
Determine whether set B is a subset of A.<br> 7. A = {2, 3, 5, 7, 11}<br> B = {3, 5, 7, 9, 11)
VARVARA [1.3K]

Answer:

Set  B is not a subset of A.

Step-by-step explanation:

If there are sets, A and B with some elements, then B can be called subset of A, if every element of B is present in set A.

Example.

Example

A is set of integers

B is set of natural numbers

we know set of integers is all the negative numbers, 0 and all the positive discrete numbers like

A = {......-3,-2,-1,0,1,2,3,4........}

B is set of positive discrete numbers from 1 to infinity)

B =  {1,2,3,4........}

Thus, we see every element of B is present in A . Hence B is subset of A

___________________________________________

Given

A = {2, 3, 5, 7, 11}

B = {3, 5, 7, 9, 11)

Here,

In set B . elements 3,5,7,11 are present in set A, but

Element 9 from set B is missing Set A.

Since, every element of Set B is not present in set A hence

Set  B is not a subset of A.

7 0
3 years ago
the scores on a psychology exam were normally distributed with a mean of 56 and a standard deviation of 9. A failing grade on th
bija089 [108]

hey! i got a failing grade! this world is so small.

3 0
3 years ago
Coffee beans worth $ 10.50 $10.50 per pound are to be mixed with coffee beans worth $ 13.00 $13.00 per pound to make 50 50 pound
MAVERICK [17]

Answer: 30 pounds of coffee beans worth $10.50 per pound was used.

20 pounds of coffee beans worth $13.00 per pound was used.

Step-by-step explanation:

Let x represent the number of coffee beans worth $10.50 per pound that should be mixed.

Let y represent the number of coffee beans worth $13.00 per pound that should be mixed.

50 pounds of beans worth $11.50 per pound is to be made. The total cost of the mixture would be

50 × 11.5 = 575

This means that

10.5x + 13y = 575 - - - - - - - 1

The total number of pounds of each type of coffee used is 50. This means that

x + y = 50

Substituting x = 50 - y into equation 1, it becomes.

10.5(50 - y) + 13y = 575

525 - 10.5y + 13y = 575

- 10.5y + 13y = 575 - 525

2.5y = 50

y = 50/2.5 = 20

x = 50 - 20= 30

6 0
4 years ago
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