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2 years ago

Help me answer please

Mathematics

1 answer:

Nastasia [14]2 years ago

5 0

Answer:

Sorry

Step-by-step explanation:

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7 divided by 2-4.5 multiplied by 3+8

nordsb [41]

Answer:

-2

Step-by-step explanation:

(7/2)-(4.5* 3) +8= -2

4 0

3 years ago

Read 2 more answers

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$

$\text{In degrees,}~ x = 135^{\circ},~315^{\circ},~\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]$

4 0

2 years ago

Can someone help with this question?

Bas_tet [7]

Get the app called connects! It’s really good help for math!

5 0

3 years ago

7. Quadrilateral MNPQ has vertices M(4,0), N(0,6), P(-4,0) and Q(0, -6). Determine whether

kirill [66]

Answer:

SQUARE

Step-by-step explanation:

If Quadrilateral MNPQ has vertices M(4,0), N(0,6), P(-4,0) and Q(0, -6).

Find the following MN, NP, PQ and MQ

Using the formula for calculating the distance between two points

MN = √(6-0)²+(0-4)²

MN = √6²+4²

MN = √36+16

MN = √52

MN = 2√13

NP = √(0-6)²+(-4-0)²

NP = √6²+4²

NP = √36+16

NP = √52

NP = 2√13

PQ = √(-6-0)²+(0-(-4))²

PQ = √6²+4²

PQ = √36+16

PQ = √52

PQ = 2√13

MQ = √(-6-0)²+(0-4)²

MQ = √6²+4²

MQ = √36+16

MQ= √52

MQ = 2√13

Since the length of all the sides are equal, hence the shape is a SQUARE

4 0

2 years ago

For f(x)=4x+1 and g(x)=x^2-5,find(f•g)(4)

RoseWind [281]

Answer:

(f•g)(4) = 45

Step-by-step explanation:

f(x)=4x+1

g(x)=x^2-5

(f•g)(x) = 4(x^2 -5)+1

(f•g)(4) = 4(4^2 -5)+1

(f•g)(4) = 4(16-5)+1

(f•g)(4) = 4(11)+1

(f•g)(4) = 44 + 1

(f•g)(4) = 45

8 0

3 years ago

Read 2 more answers