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3 years ago

Which is the graph of f(x) = -(x + 3)(x + 1)?

Mathematics

1 answer:

Bas_tet [7]3 years ago

4 0

Answer:

See attachment

Step-by-step explanation:

A function is given to us and we need to tell which graph represents the given function. The function given to us is,

$\tt: \implies f(x) = -( x + 3 )( x + 1 )$

Let's find out at which points do the graph Intersects x axis / finding the roots. For that substitute f(x) = 0 , we have ,

$\tt: \implies -(x +3)( x + 1 ) = 0$

Equate each factor by 0 ,

$\tt: \implies \boxed{\blue{ \tt x = -1,-3 }}$

Therefore the graph will intersect x axis at x is equal to -1 and x is equal to -3 .

On looking at the given graphs in the options the second graph intersects x axis at -1 and -3 .

Hence the second option is correct . 

{ See attachment }

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Which of the following is not equivalent to the expression below?

il63 [147K]

Answer: I can't see the problem

Step-by-step explanation:

3 0

3 years ago

Why does multiple of 3 add up digits trick work?

kherson [118]

You must be referring to the quick way to determine whether a given integer is divisible by 3; that is, 3 divides an integer $x$ whenever the digits of $x$ sum to a multiple of 3.

Suppose $x$ has $n\ge1$ digits. We can expand it as a sum of the numbers in any given digits place by the corresponding power of 10. For example, if $x=2148$ , we can write

$2148=2\times10^3+1\times10^2+4\times10^1+8\times10^0$

More generally, if

$x=d_{n-1}d_{n-2}\ldots d_1d_0$

(where $d_i$ denotes the numeral in the $10^i$ -th's place), then we have the expansion

$x=d_{n-1}10^{n-1}+d_{n-2}10^{n-2}+\cdots+d_110^1+d_0$

Notice that for any integer $k$ , we have

$10^k-1=\underbrace{99\ldots99}_{k\text{ 9s}}$

which is clearly divisible by 3. So from each power of 10 in the expansion of $x$ , we can add and subtract 1, then rearrange the terms of the sum:

$x=d_{n-1}10^{n-1}+\cdots+d_110^1+d_0$
$x=d_{n-1}(10^{n-1}-1+1)+\cdots+d_1(10^1-1+1)+d_0$
$x=d_{n-1}(10^{n-1}-1)+\cdots+d_1(10^1-1)+(d_{n-1}+\cdots+d_1+d_0)$

We know $10^k-1$ is divisible by 3, which means the remainder upon dividing $x$ by 3 is just the sum of the digits of $x$ . If this remainder is divisible by 3, then so must be the original number, $x$ .

Back to our previous example: if $x=2148$ , then we have the expansion

$2148=2\times10^3+10^2+4\times10+8$
$2148=2(999+1)+(99+1)+4(9+1)+8$
$2148=2\times999+99+4\times9+(2+1+4+8)$

Dividing through by 3, we get a remainder of $2+1+4+8=15$ , which is divisible by 3, and so 2148 must also be a multiple of 3.

In case for some reason you're not convinced 15 is a multiple of 3, you can apply the same trick:

$15=10+5$
$15=9+1+5$

Dividing through by 3 leaves a remainder of $1+5=6$ , which is also a multiple of 3, so that 15 must be, too.

7 0

3 years ago

Will someone plzz help

lbvjy [14]

Answer:

4.4 days

Step-by-step explanation:

375 x 0.14 = 52.5

300 - 52.5 = 247.5

247.5 / 55.98 = 4.42 days

3 0

3 years ago

HELP ME Write and solve a proportion to find y.

Phoenix [80]

Given:

A figure of dilation and the center of dilation is O.

To find:

The value of y.

Solution:

The distance between center and vertex of image (d') is proportional to the distance between center and the corresponding vertex of original figure (d).

$d'\propto d$