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3 years ago

PLEASE I NEED THE ANSWER DON'T IGNORE MY QUESTION!!!

Mathematics

2 answers:

kenny6666 [7]3 years ago

3 0

Answer:

x=7.8

Step-by-step explanation:

take 49 degree as reference angle

using cos rule

cos 49=adjacent/hypotenuse

0.65=x/12

0.65*12=x

7.8=x

pentagon [3]3 years ago

3 0

Answer:

x=7.8

Step-by-step explanation:

Step-by-step explanation:

take 49 degree as reference angle

using cos rule

cos 49=adjacent/hypotenuse

0.65=x/12

0.65*12=x

7.8=x

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11.50 + 1.15x = 12.65

Illusion [34]

Answer: the answer is x= 1

Step-by-step explanation:

So we are dealing with 11.50 - 11.50 + 1.15x = 12.65 - 11.50

our first step is Simplify both sides of the equation so:

11.5 + -11.50 + 1.15x = 12.65 + -11.50 turns into:

(1.15x) + (11.5+ -11.5)= (12.65 + -11.5)

**we know 11.5 + -11.5 will cancel each other since they are opposite**

so it will be 1.15x= 1.15

now the very last step is to divide 1.15x on both sides which gives us 1

7 0

3 years ago

Read 2 more answers

A cheerleading team practices each week for 375 minutes. The team practiced for 26 weeks during the year. How many minutes did t

Keith_Richards [23]

Answer:

9750 min or 162.5 hours

Step-by-step explanation:

375 times 26 = 9750

3 0

3 years ago

Read 2 more answers

Help I need it now!! T14= 46, and t20= 100, find t3, t7, and tn

Vinil7 [7]

Answer:

$t_3 = -53\\\\\\t_7 = -17\\t_n= -80 +9n$

Step-by-step explanation:

Arithmetic sequence:

$\sf \boxed{t_n=a+(n-1)d}$

Here a is the first term and d is the common difference.

$t_{14}= 6$ ⇒ $\sf a + 13 d = 46$ ------------(I)

$\sf t_{20} = 100$ ⇒ a + 19d = 100 ---------(II)

Subtract equation (I) from (II)

(I1) a + 19d = 100

(II) a + 13d = 46

- - -

6d = 54

d = 54 ÷ 6

$\sf \boxed{d = 9}$

Substitute d = 9 in equation(I) and find 'a',

a + 13*9= 46

a + 117 = 46

a = 46 - 117

a = -71

$\sf t_3 = -71 + 2*9$

= -71 + 18

= -53

$\sf t_7 = -71 + 6*9$

= -71 + 54

= -17

$\sf t_n= -71 + (n-1)*9$

= -71 + 9*n - 1 *9

= -71 + 9n - 9

= -71 - 9 + 9n

= - 80 + 9n

5 0

1 year ago

Can you guy help me please

nordsb [41]

For number 25 it’s x=4 and for 26 it’s x=7

7 0

3 years ago

Find parametric equations for the line through point P(-4,5,2) in the direction of the vector (-6,6,-5)

tigry1 [53]

The vector v (-6, 6, -5) points in the direction of itself, so we start there.

We can capture all points on the line through the origin and the point (-6, 6, -5) by scaling v by an arbitrary real number t.

The line through point P and pointing in the same direction as v is parallel to the other line that passes through the origin. Then the line we want can be obtained by translating the line through the origin by a vector p that points to (-4, 5, 2), so the vector equation for this line is

r (t ) = p + t v

r (t ) = (-4, 5, 2) + t (-6, 6, -5)

r (t ) = (-4 - 6t, 5 + 6t, 2 - 5t )

To get the parametric equations, simply take out the components:

x (t ) = -4 - 6t

y (t ) = 5 + 6t

z (t ) = 2 - 5t

4 0

3 years ago