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Zina [86]
3 years ago
6

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

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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}}}

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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
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