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2 years ago

5. For which linear function below does f(-1) = 9 and f(3) = 1?

Mathematics

1 answer:

Tju [1.3M]2 years ago

3 0

The answer is Y = -2X + 7

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Factorise the following :<br>

timurjin [86]

Answer:

x³+2x²+5x+10

x²(x+2)+5(x+2)

taking common

(x+2)(x²+5) is your answer

a³-a²b²-ab+b³

taking common

a²(a-b²)-b(a-b²)

taking common

(a-b²)(a²-b)

5 0

2 years ago

Read 2 more answers

Answer below please!

poizon [28]

Answer:

$x=-\frac{11}{18}$

Step-by-step explanation:

$-1\frac{7}{9}=(-2\frac{5}{6} + \frac{1}{3})x$

Solve parenthesis first

$-2\frac{5}{6} + 1/3$

Find a common denominator.

The common denominator would be 6.

Because the first expression already has a denominator of 6 it stays the same.

$\frac{1}{3} = \frac{2}{6}$

$-2\frac{5}{6} + \frac{2}{6} = -2\frac{7}{6} = -1\frac{1}{6}$

Put that into the original equation.

$-1\frac{7}{9}=(-1\frac{1}{6} )x$

Isolate the x. Add $1\frac{1}{6}$ to each side.

$-1\frac{7}{9}+ 1\frac{1}{6}=x$

Find a common denominator.

we can use 18 as the common denominator.

$-1\frac{7}{9}+ 1\frac{1}{6}= -1 \frac{14}{18} +1\frac{3}{18}= x$

Add.

$-\frac{11}{18} =x$

We cannot simplify so our answer stays the same.

6 0

1 year ago

1. One complete sentence about rotations.

Sedbober [7]

Answer: The Earth makes one rotation every day.

Step-by-step explanation:

5 0

3 years ago

Please help if you can ik it is Pythagorean theorem

kicyunya [14]

Step-by-step explanation:

A = (1/2)bh ---> h = 2A/b = 2(12 cm^2)/(5 cm) = 4.80 cm

---> x^2 = h^2 + (b/2)^2

= (4.8 cm)^2 + (2.5)^2

= 23.04 cm^2 + 6.25 cm^2

or x = 5.41 cm

Therefore, the perimeter P is

P = 2x + b = 2(5.41 cm) + 5 cm = 15.8 cm

5 0

2 years ago

If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"

seraphim [82]

Answer:

$a + b + c = \pi \\ = > c= \pi - a - b \\ < = > \tan(c) = \tan(\pi - a - b) = -\tan(a + b)$

Step-by-step explanation:

we have:

$\tan(a) + \tan(b) + \tan(c) \\ = \tan(a) + \tan(b) - \tan(a + b) \\ = \tan( a) + \tan(b) - \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{ ( \tan(a) + \tan(b) ) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (1)$

we also have:

$\tan(a) \tan(b) \tan(c) \\ = - \tan(a) \tan(b) \tan(a + b) \\ = \frac{ -(\tan( a ) + \tan(b) ) \tan(a) \tan(b) }{1 - \tan(a) \tan(b) } \\ = \frac{( \tan(a) + \tan(b)) \tan(a) \tan(b) }{ \tan(a) \tan(b) - 1 } (2)$

from (1)(2) => proven

5 0

2 years ago