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maria [59]
3 years ago
5

If RG = 4X+12 and HI = 10x-15, then x =​

Mathematics
1 answer:
Zinaida [17]3 years ago
6 0
T and g are the answer because you can’t do 10x and 15
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Simplify f(x)=(x+1)(x-3) to standard form
Ivahew [28]

Answer:

x^2-2x-3

Explanation: Use FOIL

4 0
3 years ago
Starting with the standard form of an equation ax + by = c, solve this expression for y in terms of a, b, c, and x. then put the
Vika [28.1K]
Ax+by=c
a=(-by+c)/x
b=(-ax+c)/x
c=ax+by
x=(-by+c)/a
y=(-ax+c)/b. or y= -a/b x + c/b which is slope intercept form
4 0
3 years ago
Please Help. I really don’t get this concept, if you could explain it in detail it would be very appreciated
Dafna11 [192]
<h3>Answer: 24/25</h3>

=================================================

Explanation:

Sine is given to be negative, and so is tangent. This only happens in quadrant Q4

Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.

If tan(theta) < 0, then this means cos(theta) > 0

So we have y < 0 and x > 0 which places the angle somewhere in Q4.

--------------------------

Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24

So this is what your steps may look like

a^2+b^2 = c^2

7^2+b^2 = 25^2

b^2+49 = 625

b^2 = 625-49

b^2 = 576

b = sqrt(576)

b = 24

So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25

---------------------------

As an alternative, you could use the trig identity

sin^2(x) + cos^2(x) = 1

and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.

So,

sin^2(x) + cos^2(x) = 1

(-7/25)^2 + cos^2(x) = 1

(49/625) + cos^2(x) = 1

cos^2(x) = 1 - (49/625)

cos^2(x) = (625/625) - (49/625)

cos^2(x) = (625-49)/625

cos^2(x) = 576/625

cos(x) = sqrt(576/625)

cos(x) = sqrt(576)/sqrt(625)

cos(x) = 24/25

This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)

6 0
3 years ago
Please please please please help
Andrej [43]

Answer:

true

Step-by-step explanation:

yes the statement is correct

8 0
3 years ago
All the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxe
AveGali [126]

Answer:

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

Step-by-step explanation:

Given that all the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over.

This implies that boxes are in multiples of 12

i.e. 12, or 24, or 36 or....

Let us say this as 12m for m an integer

Now additional boxes arrived =60

Total number of boxes now = 12m+60

This when arranged in stacks of 14,nothing left out

Or this is multiple of 14

12m+60 = 14n for some positive integer n.

By trial and error we find that if 60 is to be divided by 14, 24 should be added and 24 is a multiple of 12.

So we say originally 24 boxes were there now 84 boxes there.

Hence both options are right.

(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.

3 0
3 years ago
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