All categories

3 years ago

Jordan has $55 . she ears $67 by doing chores how much money does jordan have now?

1 answer:

5 0

She now has $122. If you add $55 and $67, since that is how much she earned doing chores, it will give you your answer. Hope this helped!

You might be interested in

Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k. k=______.

LenKa [72]

Answer:

Step-by-step explanation:

Look at one of the coordinates to find k. For this example I'll look at the y-intercepts (0, 3) (for g(x)) and (0, -2) (for f(x))

g(0) = 3

g(0) = f(0) + k

3 = -2 + k

5 = k

7 0

2 years ago

|×|=× is the absolute value of x always, sometimes, or never x

il63 [147K]

Answer:

Sometimes

Step-by-step explanation:

I’ll give you an example so you can understand:

Let’s say x is 4. So plug 4 into the problem:

|4|=4 → This is a very true statement, where the absolute value of 4 is equal to 4.

Now, let’s say x is -7. So plug -7 into the problem:

|-7|=-7 → This is a false statement because it’s saying that the absolute value of -7 is -7 which is very untrue.

So |x|=x only works for positive numbers, but not negative numbers. Therefore, |x|=x is the absolute value of x sometimes.

Hope this helps and answers your question! :)

6 0

3 years ago

One solution to a quadratic function, h, is given.

Margaret [11]

Answer:

B. The other solution to function $h$ is $-4-i\,7$ .

Step-by-step explanation:

From Algebra, we remember that quadratic functions of the form $a\cdot x^{2}+b\cdot x + c$ has solutions of the form:

$x = \frac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$ , where both are complex if and only if $\sqrt{b^{2}-4\cdot a \cdot c}< 0$ .

Hence, if one solution of the quadratic function is $-4+i\,7$ , then the other solution is $-4-i\,7$ . The correct answer is B.

7 0

3 years ago

Multiple all the coordinates by 1/5.

Andrew [12]

Answer:

(-1,2) , (2,-1), (2,-1) , (-2,1)

Step-by-step explanation:

-5 x 0.2=-1

5 x 0.2=1

10 x 0.2= 2

-10x 0.2=-2

6 0

3 years ago

Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7

Lapatulllka [165]

The first equation is $y-2x+1=0$

$y=2x-1$ (Equation 1)

The second equation is $4x^{2}+3y^{2}-2xy=7$ (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

$4x^{2}+3(2x-1)^{2}-2x(2x-1)=7$

$4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7$

by simplifying the given equation,

$12x^{2}-10x-4=0$

$6x^{2}-5x-2=0$

Using discriminant formula,

$D=b^{2}-4ac$

$D=25-4 \times 6 \times -2 = 73$

Now the formula for solution 'x' of quadratic equation is given by:

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12}$

Hence, these are the required solutions.

8 0

3 years ago