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2 years ago

. y = x2 + 3 y = x + 5

1 answer:

7 0

I assume you mean x squared in the first equation

Because both are equal to y, they are equal to each other so x + 5 = x^2 +3
If we then move everything over to one side, we get x^2 - x - 2 = 0

Then factorise it to (x-2)(x+1) = 0
And solve both parts separately

x + 1 = 0
x = -1

x-2 = 0
x = 2

Sub both values into the simplest equation in this case y=x+5
to get y = 4 and y = 7

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Explain how to solve the equation x + 6 = 24

jeka57 [31]

You need to subtract 6 from 24 because it’s the inverse operation of addition which will help you find the unknown value, x.

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4 years ago

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L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of

iragen [17]

Answer:

a) Assume that $x,y\in\ker L$ , and $\alpha$ is a scalar (a real or complex number).

First. Let us prove that $\ker L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity. Here, $0_V$ stands for the zero vector of V, and $0_W$ stands for the zero vector of W.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

$L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W$ .

Then, $\alpha x\in\ker L$ .

Third. Let us prove that $y+ x\in\ker L$ . Again, by linearity

$L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W$ .

And the statement readily follows.

b) Assume that $u$ and $v$ are in range of $L$ . Then, there exist $x,y\in V$ such that $L(x)=u$ and $L(y)=v$ .

First. Let us prove that range of $L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity.

Second. Let us prove that $\alpha u$ is on the range of $L$ .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of $L$ .

Third. Let us prove that $u+v$ is in the range of $L$ .

$u+v = L(x)+L(y) = L(x+y)=L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)= u +v$ . Thus $u +v$ is in the range of $L$ .

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

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3 years ago

Without solving, determine the number of real solutions for each quadratic equation.

masha68 [24]

Answer:

1. x^2 − 4x + 3 = 0

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

Step-by-step explanation:

1. x^2 − 4x + 3 = 0

We need to compare this function with the general equation for a quadratic formula given by:

$f(x) = ax^2 + bx + c$

On this case we see that a=1, b = -4 and c =3

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

We can rewrite the expression like this:

$2n^2 +4n +2$

On this case we see that a=2, b = 4 and c =2

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

We can rewrite the expression like this:

$2x^2 +x +5$

On this case we see that a=2, b = 1 and c =5

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

We can rewrite the expression like this:

$x^2 -9x -6$

On this case we see that a=1, b = -9 and c =-6

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions

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4 years ago

Use the rules of exponents to simplify the expressions. Place a final answer in fraction form. (-1/2)to the power of 3 *(-1/2) t

Pavlova-9 [17]

Answer:

-1/32

Step-by-step explanation:

we have

$(-\frac{1}{2})^{3} )(-\frac{1}{2})^{2})$

Remember that

The exponent "product rule" tells us that, when multiplying two powers that have the same base, you can add the exponents

$(-\frac{1}{2})^{3} )(-\frac{1}{2})^{2})=(-\frac{1}{2})^{3+2})=(-\frac{1}{2})^{5})=-\frac{1}{32}$

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3 years ago

c. Debbie and Jan are both in the 28% tax bracket. Since the interest is deductible, how much would Debbie and Jan each save in

alisha [4.7K]

Step-by-step explanation:

Let's assume Debbie paid $1700 in interest, while Jan paid $9000.

Debbie Saves = $1700* 0.28

Jan Saves = $9000 * 0.28

Debbie Saves $476 in taxes

Jan Saves $2520 in taxes

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3 years ago