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2 years ago

If anyone could please help me with this question???I know that you can get three triangles out of the shape.

Mathematics

1 answer:

dem82 [27]2 years ago

8 0

Answer:

See below.

Step-by-step explanation:

The triangles are similar because the angles marked 125 are equal and the sides either side of these angle are in the same ratio:

1.3 / 2.5 = 2.6 /5.

If we are given the length of AE we would find CF from the equation:

1.3 / 2.5 = CF / AE

CF = (1.3 * AE) / 2.5.

You might be interested in

Match each function with the corresponding function formula when h(x) = 5 - 3x and g(x) = -3 x + 5.

laila [671]

Answer:

To match each functions with the corresponding function formula when h(x) = 5 - 3x and g(x) = -3 x + 5.

1. k(x) = (3h - 5g)(x)

= = 3(5 - 3x)-5(-3 x + 5)

= = 15 - 9x + 15x - 25

K(x) = -10 + 6x

2. k(x) = (h - g)(x)

= = (5 - 3x) - (-3 x + 5)

= = 0

k(x) = 0

3. k(x) = (5g + 3h)(x)

= = 5(-3 x + 5)+3(5 - 3x)

= = -15x + 25 + 15 - 9x

k(x) = - 24x +40

4. k(x) = (3g + 5h)(x)

= = 3 (-3 x + 5) + 5(5 - 3x)

= = -9x + 15 +25 -15x

k(x) = -24x + 40

5. k(x) = (g + h)(x)

= = (-3 x + 5) + (5 - 3x)

k(x) = -6x + 10

6. k(x) = (5h - 3g)(x)

= = 5(5 - 3x) - 3(-3 x + 5)

= =25 - 15x + 9x - 15

k(x) = 10 - 6x

6 0

3 years ago

Can anyone help me!

zhenek [66]

Answer:

$\large \text{$ \sf 2^{-64}$}$

Step-by-step explanation:

Given:

$\large \text{$ \sf (256)^{-4^{\frac{3}{2}}}$}$

This reads as: 256 to the power of "-4 to the power of 3/2".

Therefore, we need to deal with the "-4 to the power of 3/2" first.

Rewrite the 4 as 2²:

$\large \text{$ \sf \implies -(2^2)^{\frac{3}{2}}$}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)}$}$

$\large \text{$ \sf \implies -2^{3}$}$

Therefore:

$\large \text{$ \sf \implies -2^3=(-2)(-2)(-2) = -8$}$

Replace "-4 to the power of 3/2" with -8 :

$\large \text{$ \sf \implies 256^{-8}$}$

Rewrite 256 as 2⁸ :

$\large \text{$ \sf \implies (2^8)^{-8}$}$

$\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies 2^{(8 \times -8)}$}$

$\large \text{$ \sf \implies 2^{-64}$}$

In one complete calculation:

$\large\begin{aligned} \sf (256)^{-4^{\frac{3}{2}}} & = \sf (256)^{-(2^2)^{\frac{3}{2}}}\\& = \sf (256)^{-2^{\left(2 \times \frac{3}{2}\right)}}\\& =\sf (256)^{-2^{3}}\\& = \sf (256)^{-8}\\& \sf = (2^8)^{-8}\\& = \sf 2^{(8 \times -8)}\\& =\sf 2^{-64}\end{aligned}$