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2 years ago

What is ─6d – 10 = 10 – 4d solved for the variable

Mathematics

2 answers:

patriot [66]2 years ago

6 0

Answer:

d = -10

Step-by-step explanation:

(d is the variable)

-6d - 10 = 10 - 4d

+4d +10

-2d = 20

-2d/-2 20/-2

d = -10

grin007 [14]2 years ago

3 0

Answer:

d = 2

Step-by-step explanation:

Simplify both sides.

6d − 10 = 10 − 4d

6d + −10 = 10 + −4d

6d − 10 = −4d + 10

6d − 10 + 4d = −4d + 10 + 4d

10d − 10 = 10

10d − 10 + 10 = 10 + 10

10d = 20

20 ÷ 10 = 2

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For an angle Θ with the point (12, −5) on its terminating side, what is the value of cosine?

Katyanochek1 [597]

Well, in order to answer this problem we need to use the the Pythagorean Theorem and the it will be like this:
cos = x / hypotenuse
cos= 12/13
I think with this you can figure the rest out. Hope this helps

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3 years ago

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an online seed supplier packages a seed mix that costs the company 20.70$ per pound the mix includes poppy seeds costing 24$ per

seraphim [82]

Answer:

The quantity of clover seeds in pounds to be added is 11.14

Step-by-step explanation:

Let

x -----> the quantity of poppy seeds in pounds

y -----> the quantity of clover seeds in pounds

we know that

$24(x)+13(y)=20.70(x+y)$ -----> equation A

$x=26\ pounds$ ----> equation B

substitute equation B in equation A and solve for y

$24(26)+13(y)=20.70(26+y)$

$624+13y=538.2+20,70y$

$20.70y-13y=624-538.2$

$7.7y=85.8$

$y=11.14\ pounds$

therefore

The quantity of clover seeds in pounds to be added is 11.14

8 0

3 years ago

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Tickets for a school play sold for $7.50 for each adult and $3.00 for each child. The total receipts for 118 tickets totaled $69

NeX [460]

X= number of adult tickets
y=number of child tickets.
we can suggest this system of equations:
x+y=118 ⇒x=118-y
7.5x+3y=696

we can solve this system of equations by substitution method:
7.5(118-y)+3y=696
885-7.5y+3y=696
-4.5y=-189
y=-189/-4.5
y=42

x=118-y
x=118-42
x=76

Answer: the number of adult tickets sold was 76.

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3 years ago

How would the factor pairs change if nancy had only 5 posters to arrange?

Romashka [77]

Answer:

Part A) Nancy can arrange the posters in 4 ways

Part B) Nancy can arrange the posters in 2 ways

Step-by-step explanation:

The complete question is

Nancy has 10 movie posters. She wants to hang them on a wall in equals rows.

A) Find all the ways that she can arrange the posters.

B) How would the factor pairs change if Nancy had only 5 posters to arrange?

Part A) Find all the ways that she can arrange the posters.

we know that

The factors of 10 are

1x10 -----> 1 row of 10 posters

2x5 -----> 2 rows of 5 posters

5x2 -----> 5 rows of 2 posters

10x1 -----> 10 rows of 1 poster

therefore

Nancy can arrange the posters in 4 ways

Part B) How would the factor pairs change if Nancy had only 5 posters to arrange?

we know that

if Nancy had only 5 posters to arrange

then

The factors of 5 are

1x5 -----> 1 row of 5 posters

5x1 -----> 5 rows of 1 poster

therefore

Nancy can arrange the posters in 2 ways

3 0

2 years ago

Please answer this correctly without making mistakes

Finger [1]

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3 years ago