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3 years ago

For the given equation, y = 4(x) - 1, what is the value of y for x = 2?

Mathematics

2 answers:

AysviL [449]3 years ago

8 0

Answer:

D. 7

Step-by-step explanation:

y= 4(x)-1

First: Let us take away the y =

4(x)-1

As in the question, it says that the value given for x is:

So we plug that in

4(2)-1

As in PEMDAS, multiplication becomes before subtraction

We now have

8-1

4*2=8

8-1=7

Therefore,

The answer is...

D. 7

Alik [6]3 years ago

3 0

Answer:

The answer is D (7).

Step-by-step explanation:

If you plug in 2 where x is, you get 4*2 -1, which is 8-1, which equals to 7.

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1. Solve triangle ABC if A = 42.3º, b = 12.9 meters, and c = 15.4 meters.

kumpel [21]

Answer:

a = 10.36m

B = 47.7º

C = 90º

Step-by-step explanation:

we know the angle of A so we do the sin of A; Spoiler: its 0.67

0.67 = a / 15.4

0.67 * 15.4 = a

a = 10.36

----------

B = 90 - 42.3 = 47.7º

6 0

2 years ago

the combined area of two squares is 45 square cm. Each side of one square is twice as long as the side of the other Square. What

Vlad1618 [11]

Answer:

L=6 cm l=3

Step-by-step explanation:

x+y=45

L=2l

Big Square Area

L*L=x

Little Square

l*l=y

Substituting

(L*L)+(l*l)=45

(2l*2l)+(l*l)=45

(4l^2)+(l^2)=45

5l^2=45

l^2=45/5

l^2=9

l=sqrt(9)

l=3

L=2(3)

L=6

4 0

2 years ago

Generate two equivalent fractions for each fraction. Use fraction tiles or number lines.

oksano4ka [1.4K]

Remember that finding common denominators is only one of the strategies for comparing fractions

5 0

2 years ago

If the equation; (3x)² + {27 × (3)^1/k - 15}x + 4 = 0 has equal roots find k

Dima020 [189]

Step-by-step explanation:

$\green{\large\underline{\sf{Solution-}}}$

Given quadratic equation is

$\rm :\longmapsto\:\rm \: {(3x)}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

can be rewritten as

$\rm :\longmapsto\:\rm \: {9x}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

Concept Used :- 

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Let's Solve this problem now!!!

On comparing with quadratic equation ax² + bx + c = 0, we get

$\red{\rm :\longmapsto\:a = 9}$

$\red{\rm :\longmapsto\:b = 27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15}$

$\red{\rm :\longmapsto\:c = 4}$

Since, Discriminant, D = 0

$\rm \implies\: {b}^{2} - 4ac = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 4 \times 4 \times 9 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 144 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = 144$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = {12}^{2}$

$\rm \implies\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = \: \pm \: 12$

Case - 1

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = - 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = - 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 3$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{9}$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{ {3}^{2} }$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{ - 2}$

$\rm \implies\:\dfrac{1}{k} = - 2$

$\bf\implies \:k \: = \: - \: \dfrac{1}{2}$

So, option (b) is Correct.

Case - 2

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 27$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 1$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{0}$

$\rm \implies\:\dfrac{1}{k} =0$

which is not possible.

7 0

2 years ago

(x + 3)2 + (x + 3) - 2 = 0

zhuklara [117]

Answer:

x=-7/3

Step-by-step explanation:

2(x+3)+(x+3)-2=0

2x+6+x+3-2=0

3x+7=0

3x=-7

x=-7/3

8 0

3 years ago

Read 2 more answers