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2 years ago

5 divided by a number minus eight.

Mathematics

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

5/x -8 Let x be the unknown number or variable then that is the equation

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PLEASE HELP ASAP!!!!!

attashe74 [19]

For this case what we must do is use the law of cosines.
We then have the following equation:
$c ^ 2 = a ^ 2 + b ^ 2 - 2 * a * b * cos (x)
$
Where,
a, b: sides of the triangle
x: angle between sides a and b.
Substituting values we have:
$c^2 = 90^2 + 75^2 - 2*90*75*cos(85)
$
Clearing the value of c we have:
$c = \sqrt{ 90^2 + 75^2 - 2*90*75*cos(85)}$
Answer:
An expression that is equivalent to how many feet the oak trees are from each other is:
$c = \sqrt{ 90^2 + 75^2 - 2*90*75*cos(85)}$

7 0

3 years ago

Convert 2000 g/l to the unit g/ml

balu736 [363]

$\frac{2 \000 g}{l} = \ ? \ \frac{g}{ml} \\ \\ 1 \ l = 1 \ 000 \ ml \\ \\\frac{2\ 000 \ g}{l} =\frac{2 \000 \ g}{1 \000\ ml} = 2.0 \ \ \frac{g}{ml}$

6 0

3 years ago

Read 2 more answers

If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?

bixtya [17]

Answer:

$T_{2} = 200$

Step-by-step explanation:

Given

Geometry Progression

$T_1 = 500$

$T_4 = 32$

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

$T_n = ar^{n-1}$

For first term: Tn = 500 and n = 1

$500 = ar^{1-1}$

$500 = ar^{0}$

$500 = a$

$a = 500$

For fought term: Tn = 32 and n = 4

$32 = ar^{4-1}$

$32 = ar^3$

Substitute 500 for a

$32 = 500 * r^3$

Make r^3 the subject

$r^3 = \frac{32}{500}$

$r^3 = 0.064$

Take cube roots

$\sqrt[3]{r^3} = \sqrt[3]{0.064}$

$r = \sqrt[3]{0.064}$

$r = 0.4$

Using: $T_n = ar^{n-1}$

$n = 2$ $r = 0.4$ and $a = 500$

$T_{2} = 500 * 0.4^{2-1}$

$T_{2} = 500 * 0.4^1$

$T_{2} = 500 * 0.4$

$T_{2} = 200$

<em>Hence, the second term is 200</em>

5 0

3 years ago

Please help me!!

ipn [44]

${\color{chartreuse}{Measure\:of\: ∠\:on\: one\: side\:of\:line\:=\:180°}}$

∠ Given are (x + 18)° , x° , 4x° and one Angle is 90° as per diagram

Let ,

1st∠ = (x + 18)°

2nd∠ = x°

3rd∠ = 4x°

so ,

$⟶\:(x \:+\: 18)°\:+\:x°\:+\: 4x°\:+\:90°\:=\:180°$

$⟶\:6x°\:+ \:18°\:+\:90°\:=\:180°$

$⟶\:6x°\:+ \:108°\:=\:180°$

$⟶\:6x°\:=\:180°\:-\:108°$

$⟶\:6x\:=\:72°$

$⟶\:x\:=\:\frac{72}{6} \\$

$⟶\:x\:=\:12°$

substituting x = 12 at the place of x .

= x° + 18°

= 12° + 18 °

1st∠ = 30°

__________________________________

= x°

2nd∠ = 12°

__________________________________

= 4x°

= 4(12)°

3rd∠ = 48°

7 0

3 years ago

Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral

Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

$l_{AB} = \sqrt{\left (6-4 \right )^{2}+\left (6-1 \right )^{2}} = \sqrt{29}$

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

$l_{BC} = \sqrt{\left (1-6 \right )^{2}+\left (4-6 \right )^{2}} = \sqrt{29}$

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

$l_{CD} = \sqrt{\left (-1-1 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}$

The length of side DA, with D = (-1, -1) and A = (1,4) gives;

$l_{DA} = \sqrt{\left (4-(-1) \right )^{2}+\left (1-(-1) \right )^{2}} = \sqrt{29}$

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore;

$Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1$

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

$Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1$

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

5 0

3 years ago