All categories

3 years ago

I need help with this question.

Mathematics

1 answer:

Valentin [98]3 years ago

5 0

Answer:

x = 7.5

Step-by-step explanation:

A scale factor of 0.75 means all the dimensions are multiplied by 0.75. When the dimension 10 is multiplied by 0.75, it becomes ...

... x = 0.75 × 10 = 7.5

You might be interested in

Assume that price is an integer variable whose value is the price (in US currency) in cents of an item. Assuming the item is pai

cestrela7 [59]

Answer:

$Change = (Paid*100) - Price$

Step-by-step explanation:

First of all we need to know the relation between 1 dollar and 1 cent in order to express everything in the units required in the question.

We know that $1 dollar = 100 cents$ , so the amount paid for the item with single dollars has to be multiplied by 100 to express its value in cents.

$Paid_{cents}=Paid_{dollars}*100$

Now that we have everything in cents, we can make the subtraction:

$Change_{cents} = Paid_{cents} -Price_{cents}$

Replacing the value in cents we get:

$Change = (Paid*100) - Price$

4 0

3 years ago

HELP ME PLEASE I NEED HELP

anyanavicka [17]

Answer:

Step-by-step explanation:

This is a narrower-than-normal absolute value graph, which is a v-shaped graph. It's pointy part, the vertex, lies at (2, -3) and it opens upwards without bounds along both the positive and negative x axes. Therefore, as x approaches negative infinity, f(x) or y (same thing) approaches positive infinity. As x approaches positive infinity, f(x) approaches positive infinity.

6 0

3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

<u>━━━━━━━━━━━━━━━━━━━━</u>

5 0

3 years ago

Read 2 more answers

What is 48 times 3 I'm giving y'all an easy chance for points answer is 144 I think

ch4aika [34]

Answer:

yup you are correct. The answer is 144 :)

Step-by-step explanation:

Also thx for the points!

3 0

3 years ago

Read 2 more answers

Please help in the question it should be if t=-6 when r =-2 its an error

Natasha2012 [34]

Answer:

r = - $\frac{12}{7}$

Step-by-step explanation:

Given that r varies inversely as t , then the equation relating them is

r = $\frac{k}{t}$ ← k is the constant of variation

To find k use the condition t = - 6 when r = - 2, then

- 2 = $\frac{k}{-6}$ ( multiply both sides by - 6 )

12 = k, thus

r = $\frac{12}{t}$ ← equation of variation

when t = - 7, then

r = $\frac{12}{-7}$ = - $\frac{12}{7}$

7 0

3 years ago