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3 years ago

2/3 + 3/4 = ????????????????

Mathematics

2 answers:

Sedaia [141]3 years ago

7 0

2/3 + 3/4 = 1 5/12

1 5/12 Is the answer to your question.

In-s [12.5K]3 years ago

7 0

2/3 + 3/4
Find common denominators.
Do this by multiplying the numerator and denominator of the first fraction by 4 and the second fraction by 3 (this will get you a common denominator of 12).

8/12 + 9/12
Add the numerators.

17/12
This is your final answer.

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Help please... I don't understand

olchik [2.2K]

Basically they want you to move the variables around so that it matches the standard form which is Ax+By=C.A,B,and C are just coefficients/constants.
so the original equation is
y=x-4
Minus x from both sides to move x to the left hand side.
-x+y=-4
so thats the final answer

5 0

3 years ago

Solve for x in the triangle. Round your answer to the nearest tenth.

Finger [1]

Answer:

9.7 units

Step-by-step explanation:

$\sin \: 48 \degree = \frac{x}{13} \\ \\ 0.7431448255 = \frac{x}{13} \\ x = 0.7431448255 \times 13 \\ x = 9.66088273 \\ x \approx9.7 \: units \\$

4 0

3 years ago

Janelle measures, to the nearest 1/8 inch, objects in her bedroom.

Karo-lina-s [1.5K]

The Answer: Tell Janelle to provide more information and perhaps a question instead of wasting your points. You’re welcome Janelle.

5 0

4 years ago

Name a point that is 20 units away from (5,-2)

Fudgin [204]

Answer:

Distance between both points is 20units.

Applying the Distance between Points Formula

d= √(y₂-y₁)²+(x₂-x₁)²

x₁= 5 y₁=–2

x₂= ? y₂= ?

20 = √ [y -(-2)]² + (x - 5)²

Taking the square of both sides to eliminate the square root on the right.

20² = ( y + 2)² + (x - 5)²

400 = y² + 4y + 4 + x² – 10x + 25.

400 = x² + y² – 10x + 4y + 29.

x² + y² – 10x + 4y = 400 – 29

x² + y² – 10x + 4y = 371--------------eqn 1.

We called that eqn 1 cause we need another equation to solve this.

To solve an Equation with 2 variables... You need 2 equation.

To solve that of 3 variables.... you need 3 equations(Basic Math Rules).

We're dealing with 2 variables in this case(x, y).

Now

We'd need to think in order to create another equation that'll satisfy the distance or point.

We know that the distance between both Points(known and unknown) is 20.

Half that distance is 10.

The distance of 10 will occur at the center of these 2 points.

So lets apply the formula for Mid point to get the coordinate of the center.

Midpoint

let p be the x position and q be the y position of the midpoint

p = (x₁ + x₂)/2. q = (y₁ + y₂)/2

Starting from the know point (5 , -2)

x₁=5 y₁=–2

p = (5 + x)/2. q = (–2 + y)/2

p = (x + 5)/2 q = (y–2)/2.

This is the coordinate of the Midpoint.

Now We know that the distance between (5, -2) and [(x + 5)/2 , (y–2)/2] is 10(midpoint).

Applying the distance between points again.

d = √(y₂ - y₁)² + (x₂ - x₁)²

x₁= 5 y₁= -2

x₂= (x + 5)/2 y₂= (y–2)/2

10 = √[(y–2)/2 -(-2)]² + [ (x+5)/2 – 5)]²

10 = √[(y–2)/2 + 2]² + [(x+5)/2 – 5)]²

Squaring both sides to remove square root on the right and also Taking the LCM in each bracket

We have

10² = [ (y + 2)/2]² + [ (x–5)/2]²

Distributing the "2" on the denominator to each individual number in the parentheses.

100 = (y/2 + 1)² + (x/2 – 5/2)²

100 = y²/4 + y + 1 + x²/4 –5x/2 + 25/4.

Multiply through by 4 to clear fractions.

400 = y² + 4y + 4 + x² – 10x + 25

Rearranging

x² + y² – 10x + 4y = 371.

Hmm...

Equation 1 and 2 Are the Same.

Something is Wrong!

Pls Recheck the Question

I do believe that One value of the unknown point should be given.

I maybe be wrong somewhere else too.

Pls correct in the comment section if you spot it.

7 0

3 years ago

If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.

Sophie [7]

There's no if about it,

$f(x)=x^3+3x^2-x-3
$

has a zero $f(1)=0$ so $x-1$ is a factor. That's the special case of the Remainder Theorem; since $f(1)=0$ we'll get a remainder of zero when we divide $f(x)$ by $x-1.$

At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover $f(-1)=0$ too, so $x+1$ is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.

So far we have

$x^3+3x^2-x-3 = (x-1)(x+1)(x-r)$

where $r$ is the zero we haven't guessed yet. Again we could divide $f(x)$ by $(x-1)(x+1)=x^2-1$ but just looking at the constant term we must have

$-3 = -1 (1)(-r) = r$

so

$x^3+3x^2-x-3 = (x-1)(x+1)(x+3)$

We check $f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark$

We usually talk about the zeros of a function and the roots of an equation; here we have a function $f(x)$ whose zeros are

$x=1, x=-1, x=-3$

8 0

3 years ago

Read 2 more answers