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3 years ago

Translate each equation into a verbal sentence 13 = 2 + 6t

Mathematics

1 answer:

Mama L [17]3 years ago

5 0

thirteen is equal to two plus six T. Hope this can help. Have a wonderful day. : )

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The students in Ms. Hill’s class are reading a 325-page book. There are about 90,000 words in the book.

NeX [460]

Answer:

225

Step-by-step explanation:

(90000÷325)×65=18000

18000÷80=225

reading rate of 225 words per minutes

5 0

2 years ago

The basic ratio for 24:56

kogti [31]

24:56 = 3:7
You can simplify it by dividing with 2

5 0

3 years ago

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X^2 - 4x + 3 how do I factor the expression completely

Ipatiy [6.2K]

Bx must be the sum of the numbers in the bracket and c must be the product. So we need to choose 2 numbers that multiply to make 3 and add to make -4.

A negative x negative = positive so
-3*-1=3
-3-1=-4

Therefore:
(x-3)(x-1)

8 0

3 years ago

Read 2 more answers

Substitute t=3 and t=5 to determine if the two expressions are equivalent.

motikmotik

Answer:

because both sides are equal when substituted by 3 & 5

4 0

2 years ago

25 POINTS AVAILABLE

myrzilka [38]

Answer:

$\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

1. We have the equation:

$(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2$

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2. We have the center (-4, 3) and the radius r = 6. Substitute:

$(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36$

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

7 0

3 years ago