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3 years ago

6

Jackie has one sister. Jackie's sister is 4 years older than her. The girls' ages add up to 28. Write an equation to solve for J

ackie's age. *

1 answer:

ZanzabumX [31]3 years ago

3 0

Answer:

Jackie's age = 12 years and Her sister's age = 16 years

Step-by-step explanation:

Let the age of Jackie be x.

Her sister's age = (x+4)

The sum of ages = 28

ATQ,

x+x+4 = 28

2x+4 = 28

2x = 24

x = 12

Jackie's age = 12 years

Her sister's age = (12+4) = 16 years

Hence, this is the requiredsolution.

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The bit stream generated on <em>kilobyte per second</em> would be 2304 kilobytes per second.

<u>Given the Parameters</u> :

Number of bits per sample = 24 bits

Sampling frequency = 96000 times per second

The bit rate = Sampling frequency × Number of bits per sample

The bit rate = 24 × 96000 = 2,304,000 bits per second

<u>Converting bits to kilobytes</u> ;

<em>1 kilobytes = 1000 bits</em>

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Therefore, the bit stream generated would be 2304 kbps

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3 years ago

Leila and Jo are two of the partners in a business. Leila makes $4 in profits for every $5 that Jo makes. If Jo makes $60 profit

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Answer:

<em>Leila would make a profit of $48</em>

Step-by-step explanation:

~ Knowing that there is some sort of proportion that can be formed through Leila and Jo's profit, we could form even a proportionality to compute the profit Leila makes given jo's profit of $60 ~

<em>If is is known that Leila: Jo, 4: 5, we could form such a proportion:</em>

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3 years ago

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

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

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

<u>Concept Used :- </u>

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

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

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

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

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

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

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

<u>Case - 1</u>

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

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

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

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

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

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

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

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

<em>So, option (b) is Correct. </em>

<u>Case - 2</u>

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

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

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

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

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

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

<em>which is not possible.</em>

7 0

2 years ago