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2 years ago

Solve the system of equations using substitution. Y= 2x - 10 Y= 4x - 8

Mathematics

1 answer:

Oksana_A [137]2 years ago

8 0

Answer:

Step-by-step explanation:

4x - 8 = 2x - 10

2x - 8 = -10

2x = -2

x = -1

y = 4(-1) - 8 = -4 -8 = -12

(-1, -12)

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Determine the equation of the inverse of y=1/4x^3-2

dedylja [7]

<u>Answer</u>

y⁻¹ = ∛(4x+8)

<u>Explanation</u>

y=(1/4)x³ - 2.

To find the inverse of this equation, you first make x the subject of the formular.

y=(1/4)x³ - 2

Multiply both sides by 4;

4y = x³ - 8

Add 8 on both sides of the equation;

4y + 8 = x³

x³ = 4y + 8

Apply the cube root on both sides to get the value of x;

x = ∛(4y+8)

The inverse of y=(1/4)x³ - 2 is;

y⁻¹ = ∛(4x+8)

7 0

4 years ago

Answer ASAP, will give brainliest

Eduardwww [97]

Answer:

A) 5.4 cm

Step-by-step explanation:

Length of mid segment = (sum of parallel sides )/2

Sum of parallel sides = PQ + SR = 9 + SR

$\frac{9+SR}{2}=MN\\\\\frac{9+SR}{2}=7.2 cm\\9 +SR = 7.2*2\\\\9 + SR = 14.4\\ SR = 14.4 -9 \\SR = 5.4 cm$

4 0

3 years ago

Sayyida has a circular garden of radius 21m. She ran round the garden twice. what distance did she cover?

yawa3891 [41]

answer

C = 263.76 m

Step-by-step explanation:

C = 2×pi×r

C = 2×3.14×21

C = 131.88 m

For 2 rounds,

C = 131.88 x 2

C = 263.76 m

7 0

3 years ago

Please help me find the answer

Ksivusya [100]

Answer:

-4 1/4

Step-by-step explanation:

-7 3/4 + 3 1/2 = -4 1/4

8 0

3 years ago

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

3 years ago