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3 years ago

9

The price of a holiday was decreased by 12% to £600. What was the price before

1 answer:

Naya [18.7K]3 years ago

5 0

Answeidk i tried to read it but I dont getit

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%285y%20%2B%209%29%286y%20-%201%29" id="TexFormula1" title="(5y + 9)(6y - 1)" alt="(5y + 9)(6y

mestny [16]

Hey there :)

( 5y + 9 )( 6y - 1 )

We need to use FOIL to expand, that is
First Terms
Outer Terms
Inner Terms
Last Terms

First Outer Inner Last
( 5y )( 6y ) + ( 5y )( - 1 ) + 9 ( 6y ) + 9 ( - 1 )
30y² - 5y + 54y - 9

Combine, if any, the like-terms
30y² + 49y - 9

7 0

3 years ago

Read 2 more answers

The area of the entire figure below is 1 square unit.

castortr0y [4]

Answer:

the area of the striped rectangle= length ×breadth

8 0

3 years ago

Read 2 more answers

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

Read 2 more answers

What is the domain of the function f(x) = x+1 / x2-6x+8

Ostrovityanka [42]

For this case we have that by definition, the domain of a function, is given for all the values for which the function is defined.

We have:

$f (x) = \frac {x + 1} {x ^ 2-6x + 8}$

The given function is not defined when the denominator is equal to zero. That is to say:

$x ^ 2-6x + 8 = 0$

To find the roots we factor, we look for two numbers that when multiplied give as a result "8" and when added as a result "-6". These numbers are:

$-4-2 = -6\\-4 * -2 = 8$

Thus, the factored polynomial is:

$(x-4) (x-2) = 0$

That is to say:

$x_ {1} = 4\\x_ {2} = 2$

Makes the denominator of the function 0.

Then the domain is given by:

All real numbers, except 2 and 4.

Answer:

x |x≠2,4

7 0

3 years ago

In oM name a chord that lies on a secant

krok68 [10]

I think its EB.

Hope this helps!

6 0

4 years ago