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3 years ago

HELP!! 25 Points!!!

Mathematics

1 answer:

elena55 [62]3 years ago

7 0

There you go... a. 10

If we add x to both sides, we get rid of the x on the right side which leaves us 1/2x + 2 = 7. (1/2=0.5)

Then if we substract 2 from both sides, we will get 1/2x = 5

If half of x = 5, then the whole of x = 10.

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Which inverse operation would be used to verify the following equation? 102 ÷ 3 = 34

MA_775_DIABLO [31]

An inverse operations is the reverse of the equation. For example, since it's 102 ÷ 3 = 34, to get the inverse you would take the opposite sign of "÷" and make it "×," while changing the equal sign to make the statement true. The inverse operation used to verify this is 102 = 3 × 34

5 0

3 years ago

-1.2= y/2.4 -1.7<br> Solve for y

evablogger [386]

${\huge{\boxed{\sf{Question}}$

$-1.2= y\div2.4 -1.7$

Solve for y

${\huge{\boxed{\sf{Answer\:with\:explanation }}}$

let's put our equation first

$-1.2= y\div2.4 -1.7$

Flip the equation

$y\div2.4-1.7=-1.2$

Now follow the rule of $BOMDAS/PEMDAS$

${\huge{\boxed{\sf{BOMDAS }}}$

Brackets

Of multiplication

Division

Addition

Subtraction

${\huge{\boxed{\sf{PEMDAS}}}$

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

_________________________

${\huge{\boxed{\sf{Solve\:the\:equation}}}$

_________________________

$y\div2.4-1.7=-1.2$

${\underline{simplify}}$

$0.416667y-1.7=-1.2$

add $1.7$ to both sides

$0.416667y=0.5$

Divide both sides by $0.416667$

$y=1.2$

${\huge{\boxed{\sf{Answer}}}$

$y$ would equal $1.2$

4 0

2 years ago

Lauren says that -3.5 is greater than -3 1/3. Do you agree? Explain.

kirza4 [7]

Answer:

$-3\frac{1}{2}$

Step-by-step explanation:

-3.5 means -(3 + $\frac{1}{2}$ )

In other words, -3.5 = $-3\frac{1}{2}$

When we draw two numbers $-3\frac{1}{3}$ and $-3\frac{1}{2}$ on a number line, $-3\frac{1}{2}$ comes first from left to right.

We know that all the numbers on a number line from left to right are in increasing order.

Therefore, $-3\frac{1}{2}$ is smaller than $-3\frac{1}{3}$ .

$-3\frac{1}{2}$

Lauren is not correct.

3 0

3 years ago

The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

Bezzdna [24]

Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

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

Part 9) we have

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

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

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

The vertex is the point (1,-4)

Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

6 0

3 years ago

Evaluate the line integral c y3 ds,<br> c.x = t3, y = t, 0 â‰¤ t â‰¤ 3

Temka [501]

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt$

Take $u=1+9t^4$ so that $\mathrm du=36t^3\,\mathrm dt$ . Then

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}$

7 0

3 years ago