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3 years ago

10

Help fast plz 100 points!

2 answers:

xxMikexx [17]3 years ago

6 0

Answer:

4

Step-by-step explanation:

MArishka [77]3 years ago

3 0

Step-by-step explanationdomain and range of 1/(4-9x) is second one.

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Solve by completing the square

natima [27]

Answer:

see explanation

Step-by-step explanation:

Given

x² + 10x + 14 = - 7 ( subtract 14 from both sides )

x² + 10x = - 21

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(5)x + 25 = - 21 + 25

(x + 5)² = 4 ( take the square root of both sides )

x + 5 = ± 2 ( subtract 5 from both sides )

x = - 5 ± 2

x = - 5 - 2 = - 7, or x = - 5 + 2 = - 3

-------------------------------------------------

The equation can also be solved by factoring

x² + 10x + 14 = - 7 ( add 7 to both sides )

x² + 10x + 21 = 0 ← in standard form

(x + 7)(x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 7 = 0 ⇒ x = - 7

x + 3 = 0 ⇒ x = - 3

5 0

4 years ago

Liz and Sara are making cookies for a bake sale. The graph shows the number of cookies that Liz bakes during a 5-hour period.

Viktor [21]

Liz would end up baking more cookies than sara. So the answer would be d

6 0

4 years ago

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

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

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

5 0

4 years ago

7 increased by a number w

amid [387]

Answer:

The answer is

$w + 7$

8 0

3 years ago

Read 2 more answers

24-3x=-6x+22<br> Please answer x as a fraction

kipiarov [429]

Answer:

I believe if you have a letter with no value or the value is unknown you put it over 1

4 0

3 years ago

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