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3 years ago

Solve the following equationsin¢-cos¢=0

Mathematics

1 answer:

Step2247 [10]3 years ago

3 0

Yo sup??

sinx-cosx=0

sinx=cosx

tanx=1

therefore

x=π/4

x=π-π/4

Hope this helps

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If M is the centroid of triangle GHI, HL=45, JI=63, and KG=60, find the missing measures

AysviL [449]

Answer:

Part a) $ML=15\ units$

Part b) $HM=30\ units$

Part c) $JM=21\ units$

Part d) $MI=42\ units$

Part e) $GM=40\ units$

Part f) $MK=20\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

A centroid of a triangle is the point where the three medians of the triangle meet. A median of a triangle is a line segment from one vertex to the mid point on the opposite side of the triangle.

The centroid is located two thirds of the distance from any vertex of the triangle

Part a) Find ML

we have

$HL=45\ units$

we know that

$ML=\frac{1}{3}HL$

substitute the given value

$ML=\frac{1}{3}(45)$

$ML=15\ units$

Part b) Find HM

we have

$HL=45\ units$

we know that

$HM=\frac{2}{3}HL$

substitute the given value

$HM=\frac{2}{3}(45)$

$HM=30\ units$

Part c) Find JM

we have

$JI=63\ units$

we know that

$JM=\frac{1}{3}JI$

substitute the given value

$JM=\frac{1}{3}(63)$

$JM=21\ units$

Part d) Find MI

we have

$JI=63\ units$

we know that

$MI=\frac{2}{3}JI$

substitute the given value

$MI=\frac{2}{3}(63)$

$MI=42\ units$

Part e) Find GM

we have

$KG=60\ units$

we know that

$GM=\frac{2}{3}KG$

substitute the given value

$GM=\frac{2}{3}(60)$

$GM=40\ units$

Part f) Find MK

we have

$KG=60\ units$

we know that

$MK=\frac{1}{3}KG$

substitute the given value

$MK=\frac{1}{3}(60)$

$MK=20\ units$

6 0

3 years ago

In a shipment of onions, 12% are damaged. 1,144 onions are undamaged.

katovenus [111]

Since 12% of them are damaged, that means the percentage of them that aren't 88%.
The easiest way to do it is to divide 88 from 1,144, written as 1,144÷88.
That's 13. So, then you would multiply 13 by 100. To get the total amount of onions. That'd be 1,300.

5 0

3 years ago

What integer values satisfy both inequalities? − 2 < x < 3 − 2 ≤ x < 2

NARA [144]

Given:

The inequalities are:

$-2$

$-2\leq x$

To find:

The integer values that satisfy both inequalities.

Solution:

We have,

$-2$

$-2\leq x$

For $-2$ , the possible integer values are

$x=-1,0,1,2$ ...(i)

For $-2\leq x$ , the possible integer values are

$x=-2,-1,0,1$ ...(ii)

The common values of x in (i) and (ii) are

$x=-1,0,1$

Therefore, the integer values -1, 0 and 1 satisfy both inequalities.

6 0

3 years ago

Which of the following is a true equation? k=y/x k= x/y x= y/k

faust18 [17]

Answer:

Step-by-step explanation:

Your question is not complete.

According to what do we validate with the equation.

add more info in the question because all of them can be true without any condition given or validation.

7 0

3 years ago

Show that one zero of 8x230x27 is the square of the other.

murzikaleks [220]

Answer:

8x2-30x+27 has one root (9/4) the square of the other (3/2)

Step-by-step explanation:

8x2+30x+27 = (2*x+3)*(4*x+9) => solution = {-3/2, -4/9}

One is not the square of the other because squares are positive.

8x2-30x+27 = (2*x-3)*(4*x-9) =>

solution = {3/2, 4/9} since (4/9 = (3/2)^2, therefore the preceding question is correct.

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3 years ago

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