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Katarina [22]
3 years ago
6

Help please!! 10 points <3

Mathematics
2 answers:
nikitadnepr [17]3 years ago
8 0
X=5 exactly like that person said.
tangare [24]3 years ago
5 0

Answer:

x = 5

Step-by-step explanation:

15 / 3 =5

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What is the volume of a square prism with a base area of 25 square centimeters and a height of 6 centimeters?
bonufazy [111]

Answer:

since, v=a(square) ×h

Step-by-step explanation:

v=a(squate)cm×hcm

v=25(square)cm×6cm

v=625cm×6ccm

v=3,750cm(cube)

7 0
3 years ago
. Given ????(5, −4) and T(−8,12):
damaskus [11]

Answer:

a)y=\dfrac{13x}{16}-\dfrac{129}{16}

b)y = \dfrac{13x}{16}+ \dfrac{37}{2}

Step-by-step explanation:

Given two points: S(5,-4) and T(-8,12)

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: y = mx + c where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

m = \dfrac{y_2 - y_1}{x_2 - x_1}

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

m = \dfrac{12 - (-4)}{(-8) - 5}

m = \dfrac{-16}{13}

to find the perpendicular of this gradient: we'll use:

m_1m_2=-1

both m_1and m_2 denote slopes that are perpendicular to each other. So if m_1 = \dfrac{12 - (-4)}{(-8) - 5}, then we can solve for m_2 for the slop of ther perpendicular!

\left(\dfrac{-16}{13}\right)m_2=-1

m_2=\dfrac{13}{16}:: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope m and the points (x,y). And plug into the equation: (y - y_1) = m(x-x_1)

side note: you can also use the y = mx + c to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

(y - y_1) = m(x-x_1)

we have the slope of the perpendicular to ST i.e m=\dfrac{13}{16}

and the line should pass throught S as well, i.e (5,-4). Plugging all these values in the equation we'll get.

(y - (-4)) = \dfrac{13}{16}(x-5)

y +4 = \dfrac{13x}{16}-\dfrac{65}{16}

y = \dfrac{13x}{16}-\dfrac{65}{16}-4

y=\dfrac{13x}{16}-\dfrac{129}{16}

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for T(-8,12)

(y - y_1) = m(x-x_1)

(y -12) = \dfrac{13}{16}(x+8)

y = \dfrac{13x}{16}+ \dfrac{104}{16}+12

y = \dfrac{13x}{16}+ \dfrac{37}{2}

this is the equation of the line that is perpendicular to ST and passes through T

7 0
3 years ago
Who gets this I am not sure what I have to do
natta225 [31]
Area of Parallelogram = base x height
height = 4.5 cm
base = 4.5 cm x 2

height = 4.5 cm
base = 9 cm

9 cm x 4.5 cm = 40.50 cm^2
8 0
3 years ago
Read 2 more answers
X2 + y<br> when x = -2 and y = 3.
OlgaM077 [116]
(-2)2+3= -1 hope this helps you
6 0
3 years ago
Read 2 more answers
1. take away five from twelve times f. 2. one-half of the sum of k and six 3.x squared minus the sum of 5 4.the sum of the produ
77julia77 [94]

The expressions formed are,

  1. 12f -5
  2. (k+6)/2
  3. x²-x+5
  4. ab+3c
  5. 24xy+g

Formation of expressions in 1, 2 and 3:

In 1, twelve times f is, 12f

Taking away 5, it becomes (12f-5)

In 2, sum of k and 6 is, (k+6)

One-half of the above quantity is, (k+6)/2

In 3, sum of 5 with x is, (x+5)

Now, x squared minus the above expression indicates (x²-x+5)

Formation of expressions in 4 and 5:

In 4, product of a and b, is ab and 3 times c is 3c

Sum of the expressions evaluated in the previous statement = ab+3c

In 5, 24 times the product of x and y is, 24xy

Adding, g in the above computed expression, we get, 24xy+g

Learn more about expressions here:

brainly.com/question/17167810

#SPJ4

3 0
2 years ago
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