In triangle STU s=9 cm t=15 and m

the hypotenuse of the right angled triangle is 6cm more than twice the shortest side. if the third side is 2 cm less than the hy

Mumz [18]

Answer:

Let the base be p

Hypotenuse = 2p +6

Perpendicular = 2p + 4

By Pythagoras theoram

(2p+6)^2 = (2p+4)^2 +p^2

=> 4p^2 +36 + 24p = 4p^2 + 16 +16p +p^2

=> 36+ 24p = p^2 + 16p + 16

=> p^2 - 8p - 20 = 0

=> p^2 - 10p +2p - 20 = 0

=> p(p-10) +2(p-10) = 0

=> (p-10)(p+2) = 0

p = 10 and - 2

Length can't be negative

So,

p = 10

Base = 10

Perpendicular = 24

Hypotenuse = 26

6 0

3 years ago

Shanna is planning a trip from Dallas, Texas, to San Diego, California. She knows that the distance she will travel is

enot [183]

Answer:

4.67 inches

Step-by-step explanation:

Shanna is planning a trip from Dallas, Texas, to San Diego, California.

She will travel about = 1,400 miles

The scale of a map is 300 miles = 1 inches

The trip of 1,400 miles = $\frac{1400}{300}$

= 4.66666 inches ≈ 4.67 inches

Her trip is 4.67 inches far on the map.

3 0

3 years ago

25% of what number is 21

chubhunter [2.5K]

If we call the number "x" then:
x*25%=21
x*25/100=21
x=21*4
x=84

8 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

3 years ago

18. Without solving the radical equation VX + 5 + 9 = 0, how could you tell that it has no real solution?

sweet-ann [11.9K]

Answer:

Has two unkown variables

Step-by-step explanation:

When there is only one equation provided with two unknown variables, it is difficult to solve for the unknown variables

The most simplified version of this equation would be VX = -14

More information would be needed to solve

5 0

2 years ago