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3 years ago

1.Find the perimeter of the picture frame.

Mathematics

1 answer:

labwork [276]3 years ago

5 0

Perimeter of the frame is 60 in and perimeter of the picture is 52 in. Area of the picture is 160 in²

Step-by-step explanation:

Step 1: Find the perimeter of the frame when width = 12 in and length = 18 in

Perimeter of the frame = 2 (length + width)

= 2(18 + 12) = 2 × 30 = 60 in

Step 2: Find the perimeter of the picture inside the frame.

Length of the picture = 18 - 2 = 16 in

Width of the picture = 12 - 2 = 10 in

⇒ Perimeter of the picture = 2 (length + width)

= 2 ( 16 + 10) = 2 × 26 = 52 in

Step 3: Find the area of the picture.

Area = length × width

= 16 × 10 = 160 in²

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Divide 9y power 2 by 3<br><br>divide (8x power 2 +4x ) by 2x

olchik [2.2K]

Answer:

1=27

2=34

Step-by-step explanation:

1)9y power 2÷3

or, y=9 power 2÷3

or, y=9×9÷3

or, y=81÷3

or, y=27

2) (8x power 2+4x)÷2

=(8x×8+4x)÷2

=(64x+4x)÷2

=68x÷2

or, x=68÷2

or, x=34

4 0

3 years ago

Jay sells 10 packs of apples for 80p. He wants to make £75 at the end of the year. How many packs of apples does he need to sell

ipn [44]

Answer:

937.5 packs of apples

Step-by-step explanation:

Note that:

100p = £1

80p = x

Cross Multiply

100x = 80p × £1

x = 80/100

x = £0.8

Jay sells 10 packs of apples for 80p.

Hence:

£0.8 = 10 packs of apples

£75 = x

Cross Multiply

£0.8 × x = £75 × 10 Packs

x = £75 × 10 Packs/£0.8

x = 937.5

Therefore, to make £75 at the end of the year, he has to sell 937.5 packs of apples

3 0

2 years ago

Consider the function f(x)= 2/5x-4

Gre4nikov [31]

Answer: a) $\frac{5}{2}x+10=f^{-1}(x)=g(x)$

Step-by-step explanation:

Since we have given that

$f(x)=\frac{2}{5}x-4$

a.) Find the inverse of f(x) and name it g(x).

Let f(x) = y

So, it becomes

$y=\frac{2}{5}x-4$

Switching x to y , we get

$x=\frac{2}{5}y-4$

$5x=2y-20\\\\5x+20=2y\\\\\frac{5x+20}{2}=y\\\\\frac{5}{2}x+10=y\\\\\frac{5}{2}x+10=f^{-1}(x)=g(x)$

b) . Use composition to show that f(x) and g(x) are inverses of each other.

$\mathrm{For}\:f=\frac{2}{5}x-4\:\\\\\mathrm{substitute}\:x\:\mathrm{with}\:g\left(x\right)=\frac{5}{2}x+10\\\\=\frac{2}{5}\left(\frac{5}{2}x+10\right)-4\\\\=x$

Similarly,

$\mathrm{g\left(x\right)=\frac{5}{2}x+10,\:f\left(x\right)=\frac{2}{5}x-4,\:g\left(x\right)\circ \:f\left(x\right)}\\\\\mathrm{For}\:g=\frac{5}{2}x+10\:\mathrm{substitute}\:x\:\mathrm{with}\:f\left(x\right)=\frac{2}{5}x-4\\\\=\frac{5}{2}\left(\frac{2}{5}x-4\right)+10\\\\=x$