Ask question

Ask question

All categories

3 years ago

11

Help will give 25 brainlest points

1 answer:

Alekssandra [29.7K]3 years ago

6 0

Step-by-step explanation:

The ratio of the two shorter legs of the top triangle is

$\dfrac{5}{4} = 1.25$

The bottom triangle has the following ratio:

$\dfrac{7.5}{6} = 1.25$

Since their ratios are the same, this means that the two triangles are similar.

You might be interested in

Add. (5y - 1) + (-2y + 4) A) 3y - 5 B) 3y + 3 C) 7y + 3 D) 7y - 3

Ivan

(5y - 1) + (-2y + 4) = 5y - 1 - 2y + 4
= 5y - 2y - 1 + 4
= 3y + 3 [OPTION B]

8 0

3 years ago

Which formulas maybe used to find the area of a non right triangle?

Anestetic [448]

Traditional 1/2 * base * height
Heron’s formula (it’s quite lengthy so search it up)
Hope this helped you :)

4 0

3 years ago

100pts Help rounding!

Semmy [17]

Answer:

5

Step-by-step explanation:

5 158/559

= 5 + 158/559

= 5 + 0.28264758

= 5.28264758

= 5 (rounded to nearest whole number)

4 0

2 years ago

Read 2 more answers

kotegsom [21]

Answer:

The answer is

Step-by-step explanation:

Step by step

3 0

3 years ago

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

3 years ago