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3 years ago

Fast first gets brainliest Part A Write an equation for each line.

Mathematics

1 answer:

Troyanec [42]3 years ago

6 0

Answer:

I don't really know or get this

Step-by-step explanation:

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GEOMETRY PEOPLE PLEASE HELP ME!!!!!!!!!!!!!!!!! I WILL GIVE BRAINLIEST!

IrinaVladis [17]

$\Delta KLJ\sim\DeltaNPM$ (AAA) Corresponding angles are congruent.

Therefore, the sides of the triangles are proportional:

$\dfrac{4x-4}{20}=\dfrac{30}{25}$ cross multiply

$25(4x-4)=(20)(30)$ use distributive property

$(25)(4x)+(25)(4)=600\\\\100x+100=600$ subtract 100 from both sides

$100x=500$ divide both sides by 100

$x=5$

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

$\Delta PQR\cong\Delta STU$ (SAS)

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

Therefore we have the equation:

$3y-2=y+4\qquad|+2\\\\3y=y+6\qquad|-y\\\\2y=6\qquad|:2\\\\y=3$

The perimeter of △PQR:

$P=4+6+3y-2=8+3y$

Substitute the value of y to the expression:

$P=8+3(3)=8+9=17\ ft$

7 0

3 years ago

Given f(x) = -6x - 1 and g(x) = x3, choose the expression for fºg)(x).

Gelneren [198K]

Answer:

- 6x³ - 1

Step-by-step explanation:

Substitute x = g(x) into f(x), that is

(f ￮ g)(x)

= f(x³ )

= - 6(x³) - 1

= - 6x³ - 1

4 0

3 years ago

What is the value of 12c5? A. 95,040 B.792 C.120 D.60

nydimaria [60]

12x5 is 60 is it suppose to be 12 times 5

6 0

3 years ago

Read 2 more answers

Find the area of the polygon XYZ that has its vertices at X(–3, 6), Y(–3, 1), and Z(5, 1).

elixir [45]

Hope you can understand

6 0

3 years ago

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