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2 years ago

11

Consider the 3-digit numbers which are written by using the digits 1, 2, 3, 4, 5. Find the number of those numbers which have tw

o equal digits and one different digit.

1 answer:

sergejj [24]2 years ago

4 0

2 and 5 are the answer to your question

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Can someone please answer this question please answer it correctly and lease show work please help me I need it

shepuryov [24]

Answer:

A,C,E

Step-by-step explanation:

The chosen expressions are just rewrites of the original equation.

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4 years ago

Please help !! <br><br> Will give the brainliest !!

Wewaii [24]

Answer:

The correct option is C

Step-by-step explanation:

Alternate exterior angles have the same measure .the correct option is C

6 0

3 years ago

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salantis [7]

Answer:

2

f(x)= -x + 1

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g(x) = -x + 6x +5

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h(x)= +ax + bx -c ( its form should be like that )

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3 years ago

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SashulF [63]

Answer:

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3 years ago

Need help with this maybe question

Aleonysh [2.5K]

Answer:

the vertex is:

(2, -1)

Step-by-step explanation:

First solve the equation for the variable y

$x^2-16y-4x-12=0$

Add 16y on both sides of the equation

$16y=x^2-16y+16y-4x-12$

$16y=x^2-4x-12$

Notice that now the equation has the general form of a parabola

$ax^2 +bx +c$

In this case

$a=1\\b=-4\\c=-12$

Add $(\frac{b}{2}) ^ 2$ and subtract $(\frac{b}{2}) ^ 2$ on the right side of the equation

$(\frac{b}{2}) ^ 2=(\frac{-4}{2}) ^ 2$

$(\frac{b}{2}) ^ 2=(-2) ^ 2$

$(\frac{b}{2}) ^ 2=4$

$16y=(x^2-4x+4)-4-12$

Factor the expression that is inside the parentheses

$16y=(x-2)^2-16$

Divide both sides of the equality between 16

$\frac{16}{16}y=\frac{1}{16}(x-2)^2-\frac{16}{16}$

$y=\frac{1}{16}(x-2)^2-1$

For an equation of the form

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

the vertex is: (h, k)

In this case

$h=2\\k =-1$

the vertex is:

(2, -1)

6 0

3 years ago