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4 years ago

the sum of scott's age and greg's age is 33 years if greg's age is presented by g, scott's age represented by

Mathematics

1 answer:

lisov135 [29]4 years ago

6 0

Scott would be represented by S.

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6. Jessie sorted the coins in her bank. Shemade 7 stacks of 6 dimes and 8 stacksof 5 nickels. She then found 1 dime and1 nickel.

Leokris [45]

Lets write down the information that we know.

Jessie has 7 stacks of 6 dimes, so Jessie has a total of 6*7 dimes

Dimes=7*6=42dimes

Jessie has 42 dimes, but then she found another one so lets add one more dime.

42+1=43 dimes

Jessie has 8 stacks of 5 nickels, so Jessie has 8*5 nickels

Nickels=8*5=40nickels

Jessie has 40 nickels, but then she found another one so lets add 1 to the total number of nickels.

40+1=41nickels

Now we can answer the question.

How many dimes and nickels does jessie have in all?

nickels+dimes=
41+43=84 coins

Jessie has 84 coins in total

3 0

4 years ago

Read 2 more answers

Give the coordinates of a point on the line whose equation in point-slope form is y − (−3) = 1 4 (x − 9).

statuscvo [17]

Answer:

Below

Step-by-step explanation:

● y -(-3) = (1/4) (x-9)

Replace 1/4 by 0.25 to make it easier

● y + 3 = 0.25(x-9)

● y + 3 = 0.25x - 2.25

Add 3 to both sides

● y +3-3 = 0.25x -2.25-3

● y = 0.25x -5.25

To the coordinates of a point from this line replace x by a value.

The easiest one is 0.

● y = 0.25×0 -5.25

● y = 5.25

5.25 is 21/4

So the coordinates are (0, 21/4)

7 0

3 years ago

Please answer this fast

fomenos

Answer:

CD = 2 units

Step-by-step explanation:

in the given right triangle ΔCDE,

By applying tangent rule,

tan(∠E) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

= $\frac{\text{CD}}{\text{DE}}$

tan(30)° = $\frac{\text{CD}}{2\sqrt{3}}$

$\frac{1}{\sqrt{3}}=\frac{\text{CD}}{2\sqrt{3}}$

CD = $\frac{2\sqrt{3}}{\sqrt{3}}$

CD = 2 units

Therefore, CD = 2 unit will be the answer.

4 0

3 years ago

Answer.. below: . . .

Scorpion4ik [409]

Answer:

3/12 + 239/12 = 20.16...

2309/23 + 32/32= 101.39

34/234 + 349/234=1.64

23/23-11/23= 0.52

57/78-32/78 = 0.32

Step-by-step explanation:

Just divide both sides first and then add, so it is easier to get your answer.

My answers are all rounded or repeated.

I hope this helps!

-No one

7 0

2 years ago

Read 2 more answers

Find m to cos²x-(m²-3)sinx+2m²-3=0 have root

vova2212 [387]

Answer:

$-\sqrt{2} \le m \le \sqrt{2}$ would ensure that at least one real root exists for this equation when solving for $x$ .

Step-by-step explanation:

Apply the Pythagorean identity $1 - \sin^{2}(x) = \cos^{2}(x)$ to replace the cosine this equation with sine:

$(1 - \sin^{2}(x)) - (m^2 - 3)\, \sin(x) + 2\, m^2 - 3 = 0$ .

Multiply both sides by $(-1)$ to obtain:

$-1 + \sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 3 = 0$ .

$\sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 2 = 0$ .

If $y = \sin(x)$ , then this equation would become a quadratic equation about $y$ :

$y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ .

$a = 1$ .
$b = m^{2} - 3$ .
$c = -2\, m^{2} + 2$ .

However, $-1 \le \sin(x) \le 1$ for all real $x$ .

Hence, the value of $y$ must be between $(-1)$ and $1$ (inclusive) for the original equation to have a real root when solving for $x$ .

Determinant of this quadratic equation about $y$ :

$\begin{aligned} & b^{2} - 4\, a\, c \\ =\; & (m^{2} - 3)^{2} - 4 \cdot (-2\, m^{2} + 2) \\ =\; & m^{4} - 6\, m^{2} + 9 - (-8\, m^{2} + 8) \\ =\; & m^{4} - 6\, m^{2} + 9 + 8\, m^{2} - 8 \\ =\; & m^{4} + 2\, m^{2} + 1 \\ =\; &(m^2 + 1)^{2} \end{aligned}$ .

Hence, when solving for $y$ , the roots of $y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ in terms of $m$ would be:

$\begin{aligned}y_1 &= \frac{-b + \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) + \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) + (m^{2} + 1)}{2} = 2\end{aligned}$ .

$\begin{aligned}y_2 &= \frac{-b - \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) - \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) - (m^{2} + 1)}{2} \\ &= \frac{-2\, m^{2} + 2}{2} = -m^{2} + 1\end{aligned}$ .

Since $y = \sin(x)$ , it is necessary that $-1 \le y \le 1$ for the original solution to have a real root when solved for $x$ .

The first solution, $y_1$ , does not meet the requirements. On the other hand, simplifying $-1 \le y_2 \le 1$ , $-1 \le -m^{2} + 1 \le 1$ gives:

$-2 \le -m^{2} \le 0$ .

$0 \le m^{2} \le 2$ .

$-\sqrt{2} \le m \le \sqrt{2}$ .

In other words, solving $y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0$ for $y$ would give a real root between $-1 \le y \le 1$ if and only if $-\sqrt{2} \le m \le \sqrt{2}$ .

On the other hand, given that $y = \sin(x)$ for the $x$ in the original equation, solving that equation for $x\!$ would give a real root if and only if $-1 \le y \le 1$ .

Therefore, the original equation with $x$ as the unknown has a real root if and only if $-\sqrt{2} \le m \le \sqrt{2}$ .

4 0

3 years ago