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3 years ago

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Priya tried to solve an equation step by step. \qquad\begin{aligned} \dfrac f{0.25}&=16\\\\ \\ \dfrac{f}{0.25} \cdot0.25&amp

;=16\cdot0.25&\green{\text{Step } 1}\\\\ \\ f&=4&\blue{\text{Step } 2}\\\\ \end{aligned} 0.25 f 0.25 f ⋅0.25 f =16 =16⋅0.25 =4 Step 1 Step 2

2 answers:

Zarrin [17]3 years ago

7 0

Answer:

f=4

Step-by-step explanation:

Priya Tried to solve the equation below step-by-step:

$\qquad\begin{aligned} \dfrac f{0.25}&=16\\ \dfrac{f}{0.25} \cdot0.25&=16\cdot0.25&\green{\text{Step } 1}\\ f&=4&\blue{\text{Step } 2}\\\end{aligned}$

The steps are correct and indeed f=4.

Yuki888 [10]3 years ago

3 0

Answer:

no m for khan

Step-by-step explanation:

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kipiarov [429]

Answer:

Jumper, y = 172

Shirt, x = - 105

Step-by-step explanation:

Let :

Shirt = x

Jumper = y

5x + 4y = 163 - - (1)

3x + 2y = 29 - - - (2)

Multiply (1) by 3 and (2) by 5

15x + 12y = 489 - - - (3)

15x + 10y = 145 - - - (4)

Subtract :

12y - 10y = 489 - 145

2y = 344

y = 344 / 2

y = 172

Put y = 172 in (3)

15x + 12(172) = 489

15x + 2064 = 489

15x = 489 - 2064

15x = - 1575

x = - 105

The values ;

Jumper, y = 172

Shirt, x = - 105

We weren't supposed to obtain a negative value, kindly check if the parameters given are correct.

5 0

3 years ago

X^2+1=0<br> What are the roots of this equation?<br> Thanks!

Hitman42 [59]

X² + 1 = 0

=> (x+1)² - 2x = 0

=> x+1 = √(2x)

or x - √(2x) + 1 = 0

Now take y=√x

So, the equation changes to

y² - y√2 + 1 = 0

By quadratic formula, we get:-

y = [√2 ± √(2–4)]/2

or √x = (√2 ± i√2)/2 or (1 ± i)/√2 [by cancelling the √2 in numerator and denominator and ‘i' is a imaginary number with value √(-1)]

or x = [(1 ± i)²]/2

So roots are [(1+i)²]/2 and [(1 - i)²]/2

Thus we got two roots but in complex plane. If you put this values in the formula for formation of quadratic equation, that is x²+(a+b)x - ab where a and b are roots of the equation, you will get the equation

x² + 1 = 0 back again
So it’s x=1 or x=-1

5 0

3 years ago

Read 2 more answers

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago

Please answer my question below<br> Will give you a brainliest?

VARVARA [1.3K]

Answer:

1: 3\sqrt2 - \sqrt14

2: \sqrt21 - 2\sqrt(7)

3: 4\sqrt3 - \sqrt15

4: 2\sqrt5 - \sqrt10

5: -3 + 3\sqrt5

I already gave you these answers man

Step-by-step explanation:

7 0

3 years ago

Which descriptions from the list below accurately describe the relationship

Pachacha [2.7K]

Answer:

B is the answer

Step-by-step explanation:

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3 years ago