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Lubov Fominskaja [6]
1 year ago
11

Cooper just started a running plan where he runs 8 miles the first week and then

Mathematics
1 answer:
lozanna [386]1 year ago
8 0
He would’ve ran 286 miles in total.
You might be interested in
Urgent!!!!?
ExtremeBDS [4]

Answer:

C) 6 feet

Step-by-step explanation:

The diagram is shown in the attachment.

Using the Pythagoras Theorem,

x^2=4^2+4^2

x^2=16+16

x^2=32

We take positive square root to obtain

x=\sqrt{32}

x=4\sqrt{2}

x=5.66ft

Rounding to the nearest feet we have x=6ft

4 0
3 years ago
For the given set, first calculate the number of subsets for the set, then calculate the number of proper subsets. {18, 8, 14, 9
kicyunya [14]

Answer:

Subsets of the given set = 32

Proper subset of the given set = 31

Step-by-step explanation:

Given set is {18, 8, 14, 9, 6} having 5 elements.

We know number of subsets of a set having n elements are represented by

2^{n}

where n is the number of elements in the set.

Therefore, subsets of the given set = 2^{5}

= 32

Since original set itself is a subset of its own, is not a proper subset.

Therefore, proper subset of a set = 2^{n}-1

= 2^{5}-1

= 32 - 1

= 31

7 0
3 years ago
he coordinate grid shows the graph of four equations: A coordinate grid is shown from negative 12 to positive 12 on the x axis a
denis23 [38]

The set of equation that is known to have 5 and 0 as solutions is A and B.

<h3>How to solve for the system of equations.</h3>

From the graph that we have here in this question, we are supposed to identify the equations that have this point.

We can do this by tracing the lines in order to locate 5 and 0 on the grid.

When we trace the bine that has 5 and 0, We would find out that it is traceable to A and B.

Read more on graphs here:

brainly.com/question/10465970

#SPJ1

6 0
2 years ago
Can someone plz help me. How can you find the inequalities of 11/15 and 5/7. Next 5/9 and 7/13. Next 11/15 and 5/7. Lastly 5/9 a
LekaFEV [45]
To make this a little clearer, let's give the pairs of inequalities the same denominator:

<span>Question 1: 
</span>\frac{11}{15} ? \frac{5}{7}
First, apply the common denominator to the first fraction:
(\frac{11}{15})7 \\  \frac{11}{15} *  \frac{7}{7}  \\  \frac{11*7}{15*7}  \\  \frac{77}{105}
Do the same for the second:15( \frac{5}{7}) \\  \frac{5}{7}* \frac{15}{15} \\  \frac{5*15}{7*15}  \\  \frac{75}{105}
Nest, compare the two fractions:
\frac{77}{105} \ \textgreater \   \frac{75}{105}
Therefore:
\frac{11}{15} > \frac{5}{7}
<span>
Question Two:</span>
\frac{5}{9} ? \frac{7}{13}
Apply the common denominator to fraction one:
13( \frac{5}{9}) \\  \frac{5}{9} * \frac{13}{13}  \\  \frac{5*13}{9*13}  \\  \frac{65}{117}
Fraction two:
9(\frac{7}{13}) \\  \frac{7}{13} *  \frac{9}{9}  \\  \frac{7*9}{13*9}  \\  \frac{63}{117}
Evaluate:
\frac{65}{117} > \frac{63}{117}
Therefore:
<span>\frac{5}{9} > \frac{7}{13}
</span>
Hope this helps!
5 0
3 years ago
Using the binomial theorem , obtain the expansion of :
andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1  6  15  20  15  6  1

Decreasing powers of 1 from 1^{6} to 1^{0}

Increasing powers of 3x from (3x)^{0} to (3x)^{6}

1+3x)^{6}

= 1.1^{6}(3x)^{0} + 6.1^{5}(3x)^{1} + 15.1^{4}(3x)^{2} + 20.1^{3}(3x)^{3} + 15.1²(3x)^{4} + 6.1^{1}(3x)^{5} + 1.1^{0}(3x)^{6}

= 1 + 18x + 135x² + 540x³ + 1215x^{4} + 1458x^{5} + 729x^{6}

--------------------------------------------------------------------------------------

(1-3x)^{6}

= 1.1^{6}(-3x)^{0} + 6.1^{5}(-3x)^{1} + 15.1^{4}(-3x)^{2} + 20.1^{3}(-3x)^{3} + 15.1²(-3x)^{4} + 6.1^{1}(-3x)^{5} + 1.1^{0}(-3x)^{6}

= 1 - 18x + 135x² - 540x³ + 1215x^{4} - 1458x^{5} + 729x^{6}

----------------------------------------------------------------------------------

Collecting like terms from both expressions

(1+3x)^{6} + (1-3x)^{6}

= 2 + 270x² + 2430x^{4} + 1458x^{6}

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1  5  10  10  5  1

Decreasing powers of 1 from 1^{5} to 1^{0}

Increasing powers of 2x from (2x)^{0} to (2x)^{5}

(1+2x)^{5}

= 1.1^{5}(2x)^{0} + 5.1^{4}(2x)^{1} + 10.1^{3}(2x)^{2} + 10.1^{2}(2x)^{3} + 5.1^{1}(2x)^{4}+ 1.1^{0}(2x)^{5}

= 1 + 10x + 40x² + 80x³ + 80x^{4} + 32x^{5}

8 0
3 years ago
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