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2 years ago

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The hypotenuse of a right triangle is 17 cm long. If one of the remaining two sides is 8 cm in length, then the length of the ot

her side is

2 answers:

VashaNatasha [74]2 years ago

8 0

don't know

Mark as brainlist

mate

faust18 [17]2 years ago

6 0

We will use the Pythagorean Theorem:

${17}^{2} = {8}^{2} + {x}^{2} \\ 289 = 64 + {x}^{2} \\ {x}^{2} = 289 - 64 \\ {x}^{2} = 225 \\ x = \sqrt{225} \\ x = 15$

The length of the angle is 15cm

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Which expression is correct?"The square of the sum of a number and 5, added to 3, then squared"

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2 years ago

(7x +6) - (5x - 4) = 180

BartSMP [9]

Answer:

x= 85

Step-by-step explanation:

(7x+6)-(5x-4)=180

There is a negative 1 between the two parenthesis

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I hope this helps! :)

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3 years ago

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1 cartons of juice = $4.20

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So, he saved $3.15, compared to the total ($12.60)

To calculate the percentage, we will do a proportion:

3.15 (money saved) : 12.60 (total of 3 single cartons) = x (how much percentage) : 100 (total)

3.15 : 12.6 = x : 100

To calculate x, we will moltiplicate the ends (where we have known numbers) than divide all per 12.6

So

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3 years ago

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Find the equation of the quadratic function f whose graph is shown below.

Marianna [84]

Step-by-step explanation:

A quadratic function is a second-degree polynomial function with the general form

$f(x) \ = \ ax^{2} \ + \ bx \ + \ c$ ,

where $a$ , $b$ , and $c$ are real numbers, and $a \ \neq \ 0$ .

The standard form or the vertex form of a quadratic function is, however, a little different from the general form. To get the standard form from the general form, we need to use the "complete the square" method.

$f(x) \ = \ ax^{2} \ + \ bx \ + \ c \\ \\ \\ f(x) \ = \ a\left(x^{2} \ + \ \displaystyle\frac{b}{a}x \right) \ + \ c \\ \\ \\ f(x) \ = \ a\left[x^{2} \ + \ \displaystyle\frac{b}{a}x \ + \ \left(\displaystyle\frac{b}{2a}\right)^{2} \ - \ \left(\displaystyle\frac{b}{2a}\right)^{2} \right] \ + \ c \\ \\ \\ f(x) \ = \ a\left[x^{2} \ + \ \displaystyle\frac{b}{a}x \ + \ \left(\displaystyle\frac{b}{2a}\right)^{2}\right] \ - \ a\left(\displaystyle\frac{b}{2a}\right)^{2} \ + \ c$

$f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ a\left(\displaystyle\frac{b^{2}}{4a^{2}}\right) \\ \\ \\ f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ \displaystyle\frac{b^{2}}{4a}$

Let

$h \ = \ -\displaystyle\frac{b}{2a}$ and $k \ = \ c \ - \ \displaystyle\frac{b^{2}}{4a}$ ,

then the expression reduces into

$f(x) \ = \ a \left(x \ - \ h\right)^{2} \ + \ k$ ,

where the point (<em>h</em>, <em>k</em>) are the coordinates for the vertex of the quadratic function.

There are two different methods to approach this question. First, we consider the general form of the quadratic function, it is observed that has a y-intercept at the point $\left(0, \ 2\right)$ , so

$f(0) \ = \ -2 \\ \\ \\ f(0) \ = \ a(0)^{2} \ + \ b(0) + c \\ \\ \\ c = \ -2$ .

Additionally, it is pointed that two distinct points $(-1, \ -3)$ and $(-4, \ 6)$ lies on the quadratic graph, hence

$f(-1) \ = \ -3 \\ \\ \\ f(-1) \ = \ a(-1)^{2} \ + \ b(-1) \ -2 \\ \\ \\ \-\hspace{0.36cm} -3 \ = \ a \ - \ b \ -2 \\ \\ \\ \-\hspace{0.3} a \ - \ b \ = \ -1 \ \ \ \ \ \ $-----$ \ (1)$

and

$\-\hspace{0.18cm}f(-4) \ = \ 6 \\ \\ \\ \-\hspace{0.18cm} f(-4) \ = \ a(-4)^{2} \ + \ b(-4) \ -2 \\ \\ \\ \-\hspace{0.97cm} 6 \ = \ 16a \ - \ 4b \ -2 \\ \\ \\ \-\hspace{0.98cm} 8 \ = \ 16a \ - \ 4b \\ \\ \\ 4a \ - \ b \ = \ 2 \ \ \ \ \ \ $-----$ \ (2)$ .

Subtract equation (1) from equation (2) term-by-term,

$\-\hspace{0.72cm} (4a \ - \ b) \ - \ (a \ - \ b) \ = \ 2 \ - \ (-1) \\ \\ \\ (4a \ - \ a) \ + \ \left[-b \ - \ (-b)\right] \ = \ 2 \ + \ 1 \\ \\ \\ \-\hspace{3.8cm} 3a \ = \ 3 \\ \\ \\ \-\hspace{4cm} a \ = \ 1$

Substitute $a \ = \ 1$ into equation (1),

$1 \ - \ b \ = \ -1 \\ \\ \\ \-\hspace{0.86cm} b \ = \ 2$ .

Therefore, the equation of the quadratic function is

$f(x) \ = \ x^2 \ + \ 2x \ -2$ .

$\rule{12.5cm}{0.02cm}$

Alternatively, the vertex of the quadratic function is given as the point $(-1, \ -3)$ , substitute these coordinates into the vertex form of a quadratic function.

$f(x) = a\left(x \ + \ 1\right)^{2} \ - \ 3$ .

Substitute the point $(-4, \ 6)$ into the function above,

$f(-4) \ = \ 6 \\ \\ \\ f(-4) \ = \ a\left[(-4) \ + \ 1\right]^{2} \ - \ 3 \\ \\ \\ \-\hspace{0.75cm} 6 \ = \ a(-3)^{2} \ - \ 3 \\ \\ \\ \-\hspace{0.55cm} 9a \ = \ 9 \\ \\ \\ \-\hspace{0.75cm} a \ = \ 1$ .

Therefore, the general form of the quadratic function is

$f(x) \ = \ (x \ + \ 1)^{2} \ - \ 3 \\ \\ \\ f(x) \ = \ (x^2 \ + \ 2x \ + \ 1) \ - \ 3 \\ \\ \\ f(x) \ = \ x^2 \ + \ 2x \ - \ 2$ .

6 0

2 years ago