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3 years ago

How many 3-digit numbers can you make such that they do not have any zero digit in any place?

Mathematics

2 answers:

Andre45 [30]3 years ago

7 0

Answer:

There are 648 numbers that are 3 digits numbers such that none of the digits are repeated. For the first digit, you have 9 choices (digits 1 -9) you don't want zero as the first digit.

Tanya [424]3 years ago

5 0

EDIT: There are 727 numbers that are 3 digits numbers such that none of the digits are repeated. For the first digit, you have 9 choices (digits 1 -9) you don't want zero as the first digit.------------------------------------ 727 numbers

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Please HELPPPPPP !!

vovangra [49]

We know that

<span>the triangle DEF is translated following the rule
</span>(x,y)-----> (x+2,y+2)
in the translation the figure maintains its dimensions
so
the area of triangle DEF is equal to the area of the triangle D’E’F’

therefore

the answer is
<span>The area of triangle D’E’F’ is equal to the area of triangle DEF. </span>

6 0

4 years ago

Write an equation of a line that passes through the point (8,4) and is parallel to the line y = 4x + 2.

Rudik [331]

Answer:

y=4x-28

Step-by-step explanation:

slope is 4

y-4=4(x-8)

y-4=4x-32

y=4x-32+4

y=4x-28

the slopes of parallel lines are equal

3 0

3 years ago

The answer and how it was solved

Crazy boy [7]

If you need to answer questions like these use Socratic it helps

5 0

4 years ago

Please help me with these calculus bc questions

zhannawk [14.2K]

4. Compute the derivative.

$y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1$

Find when the gradient is 7.

$4x - 1 = 7 \implies 4x = 8 \implies x = 2$

Evaluate $y$ at this point.

$y = 2\cdot2^2-2-1 = 5$

The point we want is then (2, 5).

5. The curve crosses the $x$ -axis when $y=0$ . We have

$y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4$

Compute the derivative.

$y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}$

At the point we want, the gradient is

$\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}$

6. The curve crosses the $y$ -axis when $x=0$ . Compute the derivative.

$\dfrac{dy}{dx} = 3x^2 - 4x + 5$

When $x=0$ , the gradient is

$\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}$

7. Set $y=5$ and solve for $x$ . The curve and line meet when

$5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92$

Compute the derivative (for the curve) and evaluate it at these $x$ values.

$\dfrac{dy}{dx} = 4x + 7$

$\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}$

$\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}$

8. Compute the derivative.

$y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b$

The gradient is 8 when $x=2$ , so

$2a\cdot2 + b = 8 \implies 4a + b = 8$

and the gradient is -10 when $x=-1$ , so

$2a\cdot(-1) + b = -10 \implies -2a + b = -10$

Solve for $a$ and $b$ . Eliminating $b$ , we have

$(4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}$

so that

$4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}$ .

5 0

2 years ago

(5,4) (3,7) find the equation for A+b=C

sp2606 [1]

Answer:

Step-by-step explanation:

Slope of line through (5,4) and (3,7) = (7-4)/(3-5) = -1.5

Point-slope equation of line:

y-4 = -1.5(x-5)

Convert equation to standard form:

y-4 = -1.5x + 7.5

1.5x + y = 11.5

3x +2y = 23

3 0

3 years ago