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3 years ago

How do you solve X/14=-21

Mathematics

1 answer:

Ket [755]3 years ago

8 0

You multiply -21 by 14 to find x, and that would be -21 x 14 = -294, and that is your answer.

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A class of 24 students each bought 18 chocolate kisses for their parents. How many chocolate kisses were purchased?

vagabundo [1.1K]

24x18=432 your answer is 432

8 0

3 years ago

Read 2 more answers

What is the IQR of 11,11.5,10.5,17,14.5,18,17,19

Vika [28.1K]

Ascending order :-

10.5 , 11 , 11.5 , 14.5 , 17 , 17 , 18 , 19

Median,

={(n/2)th + (n/2 +1)th}/2

={(4)th + (5)th}/2

=(31.5)/2

=15.75

I.Q.R.,

Median of lower half,

={(n/2)th + (n/2 +1)th}/2

={(2)th + (3)th}/2

=(22.5)/2

= 11.25

Median of upper half,

={(n/2)th + (n/2 +1)th}/2

={(2)th + (3)th}/2

=35/2

=17.5

I.Q.R. = 17.5 - 11.25 = 6.25

HOPE THIS WILL HELP YOU

4 0

3 years ago

Read 2 more answers

Mrs. Lopez cut 46 cm of yarn miss Hamilton cut 22 cm less than Miss Lopez how many centimeters of yarn does Miss Hamilton cut

Tcecarenko [31]

Answer: 24 cm

Step-by-step explanation:

Mrs Lopez cut 46cm of yarn.

Miss Hamilton then cut 22cm less than what Mrs Lopez had cut.

Mrs. Hamilton must have therefore cut:

This is a case of subtraction:

= 46 - 22

= 24 cm

5 0

3 years ago

PLEASE HELP

Dmitry_Shevchenko [17]

Answer:

2.5 betul tak........

Step-by-step explanation:

i/o

8 0

3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago