Help I’ll give brainliest!

Mathematics

1 answer:

natima [27]3 years ago

8 0

Answer:

(D) -0.2

Step-by-step explanation:

slope = 7-5 / 15 - 25

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Which trigonometric ratios are correct for triangle ABC? Check all that apply

sergij07 [2.7K]

Consider an angle M with measure m≠90°, in a right triangle.

Let

OPP denote the length of the side opposite to M,
ADJ denote the length of the side adjacent to M, and
HYP denote the hypotenuse.

then:

Sin(M) = OPP/HYP
Cos(M)= ADJ/HYPP
Tan(M)=OPP/ADJ

Back to our problem,

using the Pythagorean we can find the length of AB:

$|AB|^2+|AC|^2=|BC|^2\\\\|AB|^2+9^2=18^2\\\\|AB|^2=18^2-9^2\\\\|AB|^2=(18-9)(18+9)=9 \cdot 27=9 \cdot 9 \cdot3\\\\|AB|=9 \sqrt{3} 
$

$Sin(C)= \frac{OPP}{HYP}=\frac{9 \sqrt{3} }{18}= \frac{ \sqrt{3} }{2}$
$Cos(B)= \frac{ADJ}{HYP}= \frac{9 \sqrt{3}}{18}= \frac{ \sqrt{3}}{2}$
$Tan(C)= \frac{OPP}{ADJ}= \frac{9 \sqrt{3} }{9}= \sqrt{3}$
$Sin(B)= \frac{OPP}{HYP}= \frac{9}{18}= \frac{1}{2}$
$Tan(B)= \frac{OPP}{ADJ}= \frac{9}{9 \sqrt{3}}= \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3}}{3}$

Answer: 1, 3, 4

5 0

3 years ago

20 points plus that brain thing

madam [21]

Answer:

$\huge\boxed{\sf \$ \ 26.24}$

Step-by-step explanation:

Each side of the squared painting = 41 inches

A square has 4 sides. So,

4 sides of the square painting = 4(41) inches

= 164 inches

$\rule[225]{225}{2}$

<u>The shop charges:</u>

1 inches = $ 0.16

Multiply both sides by 164

164 inches = $ 0.16 × 164

164 inches = $ 26.24

$\rule[225]{225}{2}$

Hope this helped!

<h2>~AnonymousHelper1807</h2>

5 0

3 years ago

Simplify the square root of four over the cubed root of four.

Vikki [24]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\\\ ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\sqrt{4}}{\sqrt[3]{4}}\implies \cfrac{\sqrt[2]{4}}{\sqrt[3]{4}}\implies \cfrac{4^{\frac{1}{2}}}{4^{\frac{1}{3}}}\implies 4^{\frac{1}{2}}\cdot 4^{-\frac{1}{3}}\implies 4^{\frac{1}{2}-\frac{1}{3}}\implies 4^{\frac{3-2}{6}} \\\\\\ 4^{\frac{1}{6}}\implies (2^2)^{\frac{1}{6}}\implies 2^{2\cdot \frac{1}{6}}\implies 2\frac{1}{3}\implies \sqrt[3]{2}$