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3 years ago

Write an expression that represents the products of q and 5 divided by 2

Mathematics

1 answer:

Slav-nsk [51]3 years ago

7 0

Product of q and 5= 5q (Constant is always first, followed by variable)

Product divided by 2= 5q/2 .

∴ Expression= 5q/2

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Find the missing side. Round to the nearest tenth.

Firdavs [7]

Answer: 23.7

Step-by-step explanation:

Let's calculate the acute angle at the base of the triangle:

180° - 90° - 40° = 50°

Find side x:

$\dfrac{sin \: 90^{o} }{31} =\dfrac{sin \: 50^{o} }{x}\\\\sin \: 90^{o} \cdot x=sin \: 50^{o} \cdot 31\\\\sin \: 90^{o} =1\\sin \: 50^{o} \approx 0.766\\1 \cdot x =0.766 \cdot 31\\x=23.746\\23.746 \approx 23.7$

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2 years ago

Help please! using quadratic formula round ti nearest hundred

Lera25 [3.4K]

Answer:

x = 0.19 or x = -0.76

Step-by-step explanation:

Equation: 7²-4x = 2

reordering the equation,

7x²-4x-2 = 0

Applying,

x = [-b±√(b²-4ac)]/2a..................... Equation 1

From the question,

Given: a = 7, b = -4, c = -2

Substitute these values into equation 1

x = [-4±√(-4²-(2×7×-2)]/(2×7)

x = [-4±√(16+28)]/14

x = (-4±√44)/14

Either

x = (-4+√44)/14

x = (-4+6.63)/14

x = 2.63/14

x = 0.19

x = (-4-√44)/14

x = (-4-6.63)/14

x = -10.63/14

x = -0.76

3 0

3 years ago

Please help me out :P

Veseljchak [2.6K]

cos θ = $\frac{-4\sqrt{65} }{65}$ , sin θ = $\frac{-7\sqrt{65} }{65}$ , cot θ = 4/7, sec θ = $\frac{-\sqrt{65} }{4}$ , cosec θ = $\frac{-\sqrt{65} }{7}$

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = $\sqrt{7^{2} + 4^{2} }$ = $\sqrt{49+16}$ = $\sqrt{65}$

sin θ = opposite/ hyp = $\frac{7}{\sqrt{65} }$

Rationalize, $\frac{7}{\sqrt{65} }$ x $\frac{\sqrt{65} }{\sqrt{65} }$ = $\frac{7\sqrt{65} }{65}$

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - $\frac{7\sqrt{65} }{65}$

cos θ = adj/hyp = $\frac{4}{\sqrt{65} }$

By rationalizing and knowing that cos θ is negative, cos θ = - $\frac{-4\sqrt{65} }{65}$

cot θ = 1/tan θ = 1/7/4 = 4/7

sec θ = 1/cos θ = 1/ $\frac{4}{\sqrt{65} }$ = - $\frac{-\sqrt{65} }{4}$

cosec θ = 1/sin θ = 1/ $\frac{\sqrt{65} }{7}$ = $\frac{-\sqrt{65} }{7}$

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0

1 year ago

this is 200 points my insta is niqdariot for the other 100 pleaseeeeeeeee help me asap its only 4 questions and I need all four

Kamila [148]

Answer:

19. parallel

20. corresponding

21. exterior

22. corresponding exterior angles

if not for the points i wouldve left u to suffer on ur own so yor welcome

8 0

3 years ago

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The first three steps in determining the solution set of the system of equations algebraically are shown in the table. y = −x2 +

lianna [129]

You have a system of equations $\left \{ {{y = -x^2+2x-9} \atop {y =-6x+6}} \right.$ .

1. Substitude right side of second equation into the left side of the first equation: $-6x+6=-x^2+2x-9$ .

2. Solve this equation:
$x^2-2x+9-6x+6=0, \\ x^2-8x+15=0, \\ D=(-8)^2-4\cdot 1\cdot 15=64-60=4, \\ \sqrt{D}=2, \\ x_{1,2}=\dfrac{8\pm 2}{2 } =3,5$ .

3. Find y:
for $x_1=3, y_1=-6\cdot 3+6=-18+6=-12$ ,
for $x_2=5, y_2=-6\cdot 5+6=-30+6=-24$ .

4. The solutions of the system are: (3,-12) and (5,-24).
Answer: Correct choice is A.

7 0

3 years ago

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