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Oduvanchick [21]

3 years ago

14

I cant get this one, 216−(−83)+(−479)

1 answer:

Inessa [10]3 years ago

7 0

Answer:

180

Step-by-step explanation:

=216-(-83)+(-479)

=216+83-479

=299-479

=180

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URGENT! WILL GIVE BRAINLIEST!!! IF CORRECT!

In-s [12.5K]

Answer: $6.75

Step-by-step process:
100%-25%=75%
therefore, they will cost 75% of the original price
75% of $9.00 = .75*9
= $6.75

5 0

2 years ago

Read 2 more answers

1) Graph the following lines and using the graph determine several values of x for which the graph is positive (negative):

Aleksandr [31]

1) $y=-0.5x-2$

The graph of the above equation is attached in the attachment below.

For all the values from negative infinity to -4, the function is positive.

X ∈ (-∞,-4]

2) $y=-1.5x+3$

To find the x- intercept, plug y =0

$0=-1.5x+3$

$1.5x=3$

$x=2$

x intercept = (2,0)

To find the y- intercept, plug x=0

$y=-1.5(0)+3$

$y=3$

Y- intercept = ( 0,3)

3) $y=13-x$

Y- intercept = (0,13)

x- intercept = (13,0)

Area of triangle = $\frac{bh}{2}$

Area of triangle = $\frac{13\times13}{2}$

Area = $\frac{169}{2}$ sq unit.

4) The graphs of $y=-2x+7$ and $y=0.5x-5.5$ is attached in the attachment below.

The intersection point is ( 5,-3)

5) A = (-3,0)

B = (4,5)

C = (0,-4)

Slope of AB = $\frac{5-0)}{4-(-3)} =\frac{5}{7}$

Slope intercept of line $y=mx+b$

Where, m is the slope and b is the y- intercept.

Plugging point A and the slope in the y- intercept to find the value of b.

$0=\frac{5(-3)}{7} +b$

$b=\frac{15}{7}$

Equation of line AB: $y=\frac{5x}{7} +\frac{15}{7}$

Slope of BC = $\frac{-4-5}{0-4} =\frac{9}{4}$

Plug C = (0,-4)

$-4=0+b$

b=-4

equation of line BC= $y=\frac{9x}{4} -4$

Slope of line AC = $\frac{-4-0}{0--3} =\frac{-4}{3}$

Equation of line AC = $y=\frac{-4x}{3} -4$

Y intercept of AB = $(0,\frac{15}{7} )$

5 0

3 years ago

Read 2 more answers

Help me ;v; and thank you sm!

cupoosta [38]

27 Cm² (C)

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7 0

2 years ago

Read 2 more answers

Four resistors are connected in series such that their total resistance is 50ohm . If the resistance of resistors is 5 ohm , 10o

Katyanochek1 [597]

<h2>Answer:</h2>

$\boxed{R_{4}=20\Omega}$

<h2>Step-by-step explanation:</h2>

If resistors are in series, so the current $I$ is the same in all of them. In this problem we have four resistors. So, we can get a relationship between the Equivalent resistance of series combination and the four resistors as follows:

$R_{eq}=R_{1}+R_{2}+R_{3}+R_{4}$

$R_{eq}$ is the total resistance $50\Omega$ . Moreover:

$R_{1}=5\Omega \\ R_{2}=10\Omega \\ R_{3}=15\Omega$

Therefore:

$50=5+10+15+R_{4} \\ \\ R_{4}=50-(5+10+15) \\ \\ R_{4}=50-30 \\ \\ \boxed{R_{4}=20\Omega}$

7 0

3 years ago

A standard letter-sized piece of paper has a length and width of 8.5 inches by 11 inches.

den301095 [7]

Answer:

Area of the resulting rectangle = 23.375 inch²

Step-by-step explanation:

After 1 fold width or length reduces to half.

Area = Length x width

So after 1 fold area will reduce to half.

So we have

$\texttt{Area after n folds =}\frac{\texttt{Initial area}}{2^{\texttt{number of folds}}}$

Initial area = 8.5 x 11 = 93.5 inch²

Here the paper were folded completely in half, length is reduced to half and width reduced to half, that is

Number of folds = 2

$\texttt{Area after 2 folds =}\frac{\texttt{Initial area}}{2^{2}}\\\\\texttt{Area after 2 folds =}\frac{93.5}{4}=23.375inch^2$

Area after 2 folds = 23.375 inch²

Area of the resulting rectangle = 23.375 inch²

8 0

2 years ago