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3 years ago

What is the first step in solving the following system of equations?

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer:

Solve the 1st equation for y and then you substitute the answer of the y value to the other equation to get the x value

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You wish to make a triangular garden. One of the angles is 26°. The second angle is complementary to the first angle. What is th

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Since the second angle + first angle is 90 degrees, the third angle must be 90 degrees to complete the 180 degrees in a circle.
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3 years ago

What is the range of the following relation?<br> Oxso<br> all real numbers

Vesnalui [34]

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3 years ago

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What are the zeros of the following quadratic equation: y = 6x2 - 17x - 3

Mama L [17]

Answer:

$\displaystyle x=-\frac{1}{6}, 3$

Step-by-step explanation:

Hi there!

$y = 6x^2 - 17x - 3$

Factor by grouping:

$y = 6x^2 - 18x+1x - 3\\y = 6x(x- 3)+1x - 3\\y = 6x(x- 3)+(x - 3)\\y = (6x+1)(x- 3)$

Let y=0. Apply the zero product property:

$0 = (6x+1)(x- 3)$

$6x+1=0\\6x=-1\\\\\displaystyle x=-\frac{1}{6}$

AND

$x-3=0\\x=3$

I hope this helps!

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3 years ago

Find the surface area of the following pyramid using its net. Please answer correctly the 4 questions and I'll give you brainlie

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Answer:

The surface is 7.4344433322 acording to the exact dynamiter and angle of the pyrmaids

Step-by-step explanation

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3 years ago

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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago