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3 years ago

Can anybody help me please

Mathematics

1 answer:

GREYUIT [131]3 years ago

8 0

Question 7: Option 1: x = 33.5°

Question 8: Option 3: x = 14.0°

Step-by-step explanation:

Question 7:

In the given figure, the value of perpendicular and hypotenuse is given, so we have to use any trigonometric ratio to find the value of angle as the given triangle is a right-angled triangle

So,

Perpendicular = P = 32

Hypotenuse = H = 58

So,

$sin\ x = \frac{P}{H}\\sin\ x = \frac{32}{58}\\sin\ x = 0.5517\\x = sin^{-1} ( 0.5517)\\x =33.48$

Rounding off to nearest tenth

x = 33.5°

Question 8:

In the given figure, the value of Base and Perpendicular is given, we will use tangent trigonometric ratio to find the value of x

So,

Perpendicular = P = 5

Base = B = 20

So,

$tan\ x = \frac{P}{B}\\tan\ x = \frac{5}{20}\\tan\ x = 0.25\\x = tan^{-1} (0.25)\\x = 14.036$

Rounding off to nearest tenth

x = 14.0°

Keywords: Right-angled triangle, trigonometric ratios

Learn more about trigonometric ratios at:

#LearnwithBrainly

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astra-53 [7]

By substitution is how to solve

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y=3-6 = -3

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5 0

3 years ago

Read 2 more answers

PLS HELP find the unknown side of the triangle below (round to the nearest tenth)

Nonamiya [84]

22.9 is the closest but not exact. It was probably rounded off.

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7 0

3 years ago

Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the func

mote1985 [20]

Solution:

The given Polynomial is :

$f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$

By Rational Root theorem the of Zeroes of the Polynopmial are:

$\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4$

But , $f(\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4)\neq 0$

So, no root of this polynomial is real.

Therefore, All the four roots of Polynomial are imaginary.

So, we can't say whether the number k=2, is an upper or lower bound of the polynomial $f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})$ .