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3 years ago

A coordinate plane is shown.

Mathematics

2 answers:

frutty [35]3 years ago

8 0

Answer:

The answer is C

Step-by-step explanation:

First you would start at the origin then move 3 units to the right then move 6 units up then that is your point

Inessa [10]3 years ago

7 0

The only answer that makes sense is C.

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1. Which are true of x = log10 31,500? Select all that apply.

ziro4ka [17]

Given:

$x=\log_{10}31,500$

To find:

Select the true statements from the given options about the given value.

Solution:

We have,

$x=\log_{10}31,500$

It can be written as

$x=\log_{10}(2^23^25^37)$

$x=\log_{10}(2^2)+\log_{10}(3^2)+\log_{10}(5^3)+\log_{10}(7)$ $[\because \log(ab)=\log a+\log b]$

$x=2\log_{10}2+2\log_{10}3+3\log_{10}5+\log_{10}(7)$

$x=2\left(0.30105\right)+2\left(0.47712\right)+3\left(0.69897\right)+0.8451$

$x=4.49835$

Clearly, the value of x lies between 4 and 5. So, $x>4$ and $x$ .

Therefore, the correct options are C and D.

4 0

3 years ago

(a) Estimate the number of boys at Hillview.

3241004551 [841]

I say around 210-220 is a good answer

4 0

4 years ago

Which best explains whether a triangle with side lengths 2in , 5in , 4in. is an acute triangle

lutik1710 [3]

So we know that 3,4 and 5 are pythagorean triplet and these result from a right triangle sides length
- than 3,4 and 5 mean right triangle so 2,5 and 4 mean acute triangle sure or you can prove it in this way too
- than 3^2 +4^2 =5^2 => a right triangle so than
2^2 +4^2 < 5^2 => an acute triangle

hope helped

3 0

3 years ago

THIS IS MULTIPLE CHOICE!

Troyanec [42]

Answer:

there is a cluster from 1-4

there is a gap from 5-7

The spread is from 1-8

Step-by-step explanation:

These are all true if you look... there is a bunch from 1-4, there is none from 5-7, and all of them are within 1-8, but the highest amount isn't at 3.

7 0

3 years ago

What is the y-value of the vertex of the function f(x)=-(x-3)(x+11)

lutik1710 [3]

so, this is a quadratic equation, meaning two solutions, and we have a factored form of it, meaning you can get the solutions by simply zeroing out the f(x).

$\bf \stackrel{f(x)}{0}=-(x-3)(x+11)\implies 0=(x-3)(x+11)\implies x= \begin{cases} 3\\ -11 \end{cases} \\\\\\ \boxed{-11}\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}-4\stackrel{\textit{\large 7 units}}{\rule[0.35em]{10em}{0.25pt}}\boxed{3}$

so the zeros/solutions are at x = 3 and x = -11, now, bearing in mind the vertex will be half-way between those two, checking the number line, that midpoint will be at x = -4, so the vertex is right there, well, what's f(x) when x = -4?

$\bf f(-4)=-(-4-3)(-4+11)\implies f(-4)=7(7)\implies f(-4)=49 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{vertex}{(-4~~,~~49)}~\hfill$

7 0

3 years ago