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2 years ago

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Angela tried to solve an equation step by step. \begin{aligned} \dfrac34+m&=\dfrac54\\\\ \dfrac34+m-\dfrac34&=\dfrac54+\

dfrac34&\green{\text{Step } 1}\\\\ m&=2&\blue{\text{Step } 2} \end{aligned} 4 3 +m 4 3 +m− 4 3 m = 4 5 = 4 5 + 4 3 =2 Step 1 Step 2 Find Angela's mistake.

1 answer:

Liono4ka [1.6K]2 years ago

7 0

Answer:

The mistake she made was ; She didn't change the sign when subtracting -3/4 from both sides

She wrote 5/4+3/4 instead of 5/4-3/4

Step-by-step explanation:

Correct solution:

$\frac{3}{4} + m = \frac{5}{4}$

Subtract-3/4 from both sides

$\frac{3}{4} - \frac{3}{4} + m = \frac{5}{4} - \frac{3}{4}$

Simplify

$m = \frac{5}{4} - \frac{3}{4} \\ m = \frac{1}{2}$

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The weight Wkg of a metal bar varies jointly

maria [59]

Answer:

d = $\sqrt{\frac{216W}{35L} }$

Step-by-step explanation:

Given that W varies jointly as L and d² then the equation relating them is

W = kLd² ← k is the constant of variation

To find k use the condition W = 140 when d = 4 and L = 54, thus

140 = k × 54 × 4² = 864k ( divide both sides by 864 )

$\frac{140}{864}$ = k , that is

k = $\frac{35}{216}$

W = $\frac{35}{216}$ Ld² ← equation of variation

Multiply both sides by 216

216W = 35Ld² ( divide both sides by 35L )

$\frac{216W}{35L}$ = d² ( take the square root of both sides )

d = $\sqrt{\frac{216W}{35L} }$

6 0

2 years ago

Seven customers enter a clothing store and are observed to determine if they will make a purchase. The store owner states that t

blondinia [14]

Answer: Mean = 5.25

Variance = 1.3125

Standard deviation = 1.456

Step-by-step explanation:

The mean and variance of a binomial distribution with parameters n (number of trials) and p ( the probability of getting success in each trial) is given by :-

$\text{Mean}=np\\\\\text{Variance}=np(1-p)$

Given : The total number of enter a clothing store : n=7

The probability of a customer making a purchase= 0.75

Now,

$\text{Mean}=7(0.75)=5.25\\\\\text{Variance}=7(0.75)(0.25)=1.3125\\\\\Rightarrow\ \text{Standard deviation}=\sqrt{1.3125}=1.14564392374\approx1.1456$

3 0

2 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

2 years ago

Help me fill in the blanks<br><br> 6x + __ =3<br> (__ + 5)

Virty [35]

Answer:

<u>15</u>

<u>2x</u>

Step-by-step explanation:

4 0

2 years ago

The altitude (i.e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing

Stolb23 [73]

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2 years ago