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Ilya [14]
3 years ago
15

PLEASE HELP!! MATH QUESTION!! 20 points!!

Mathematics
2 answers:
Olenka [21]3 years ago
7 0

Answer:

the first one is a posituve

and the last 2 are negative

Step-by-step explanation:

Blababa [14]3 years ago
5 0
The first one is a posituve 

and the last 2 are negative



hope this help
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What are the multiples of 78
Yuki888 [10]
The multiples are: 78,156,234,312,390,468,546,624,702,780.
7 0
4 years ago
Read 2 more answers
If line L in the xy-coordinate plane has a positive slope, what is the x-intercept of L ? (1) There are different points (a, b)
umka2103 [35]

Answer:

The x-intercept of L is 0

Step-by-step explanation:

1)  If (a,b) and (c,d) belongs to theline L, then the equation of the line using two-points formula  will be:

y-b=(\frac{b-d}{a-c} ).(x-a)

if we want to find the x-intercept of L we should set y=0.

0-b=(\frac{b-d}{a-c} ).(x-a)

getting x from that equation we will have :

x=-b.\frac{a-c}{b-d}+a

using distributive propertie and common denominator will be obtain

x= x= \frac{bc-ad}{b-d}

as we know that ad=bc the numerator will be equal to zero. Then x=0.

2) Using the same equation of line but using the points(m,n) and (-m, -n) we will set it as:

y-n=\frac{-2n}{-2m}.(x-m)

if we want to find the x-intercept of L we should set y=0.

0-n=\frac{-2n}{-2m}.(x-m)

getting x from that equation we will have :

-n=\frac{n}{m}(x-m)

x= -n.\frac{m}{n} + m

them

x = 0

4 0
3 years ago
Find the dimensions of a rectangle (in m) with area 1,728 m2 whose perimeter is as small as possible. (Enter the dimensions as a
lozanna [386]

Given :

Area of rectangle.

To Find :

The dimensions of a rectangle (in m) with area 1,728 m2 whose perimeter is as small as possible.

Solution :

Let, the dimensions of rectangle is x and y.

Area, A = xy.

x = A/y.              ....1)

Perimeter, P = 2( x + y )

Putting value of x in above equation, we get :

P = 2( y + \dfrac{A}{y})

For minimum P,

\dfrac{dP}{dx}=2( 1 - \dfrac{A}{x^2})=0\\\\A = x^2

SO, it is a square.

x=\sqrt{A}\\\\x=\sqrt{1728\ m^2}\\\\x=24\sqrt{3}\ m

Therefore, the dimensions are (24\sqrt{3},24\sqrt{3}).24\sqrt{3}

Hence, this is the required solution.

7 0
3 years ago
PLEASE HELP WITH THIS ONE QUESTION
Furkat [3]

Answer:

(x-1) (x-4) is the answer

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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.
Mademuasel [1]

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