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Makovka662 [10]

2 years ago

12

If the measure of angle ACB is 90°, then which expression represents the value of g? triangle ACB, point E is on segment AC betw

een points A and C and point D is on segment BC between points B and C, creating segment ED, CE equals h, EA equals j, CD equals k, DB equals m, ED equals g, and AB equals f
g equals f over 2
g = 2f
g equals j over h
g equals k over m

1 answer:

Ulleksa [173]2 years ago

3 0

Answer:

$g=\frac{f}{2}$

Step-by-step explanation:

The picture of the question in the attached figure

In this problem i will assume that point E is the midpoint AC and point D is the midpoint BD

so

we know that

The <u><em>Triangle Midpoint Theorem</em></u> states that the segment joining two sides of a triangle at the midpoints of those sides, is parallel to the third side and is half the length of the third side

In this problem the segment ED joining two sides of a triangle at the midpoints of those sides

so

ED is parallel to AB and ED is half the length of AB

$ED=\frac{1}{2}AB$

we have

$AB=f\\ED=g$

substitute the given values

$g=\frac{f}{2}$

Guest

1 year ago

so whats the answer

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Jane is taking two courses. The probablity she passes the first course is 0,7. The probablity she passes the second course is 0.

makkiz [27]

Answer:

a. 0.6

b. not independent

c. 0.1

d. 0.4

e. 0.3

Step-by-step explanation:

a.

P(passing first course)=P(C1)=0.7

P(passing second course)=P(C2)=0.8

P(passing at least one course)=P(C1∪C2)=0.9

P( passes both courses)=P(C1∩C2)=?

We know that

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∩B)=P(A)+P(B)-P(A∪B)

So,

P( passes both courses)=P(C1∩C2)=P(C1)+P(C2)-P(C1∪C2)

P( passes both courses)=P(C1∩C2)=0.7+0.8-0.9

P( passes both courses)=P(C1∩C2)=0.6

Thus, the probability she passes both courses is 0.6.

b.

The event of passing one course is independent of passing another course if

P(C1∩C2)=P(C1)*P(C2)

P(C1)*P(C2)=0.7*0.8=0.56

P(C1∩C2)=0.6

As,

0.6≠0.56

P(C1∩C2)≠P(C1)*P(C2),

So, the event of passing one course is dependent of passing another course.

c.

P(not passing either course)=P(C1∪C2)'=1-P(C1∪C2)

P(not passing either course)=P(C1∪C2)'=1-0.9

P(not passing either course)=P(C1∪C2)'=0.1

Thus, the probability of not passing either course is 0.1.

d.

P(not passing both courses)=P(C1∩C2)'=1-P(C1∩C2)

P(not passing both courses)=P(C1∩C2)'=1-0.6

P(not passing both courses)=P(C1∩C2)'=0.4

Thus, the probability of not passing both courses is 0.4.

e.

P(passing exactly one course)=?

P(passing exactly course 1)=P(C1)-P(C1∩C2)=0.7-0.6=0.1

P(passing exactly course 2)=P(C2)-P(C1∩C2)=0.8-0.6=0.2

P(passing exactly one course)=P(passing exactly course 1)+P(passing exactly course 2)

P(passing exactly one course)=0.1+0.2

P(passing exactly one course)=0.3

Thus, the probability of passing exactly one course is 0.3.

3 0

2 years ago

Kx + 3x = 4 solve for x.

victus00 [196]

Take out a common factor between 3x and kx. That means use the distributive law to get what you normally would start with.

x(k + 3) = 4

Now divide by k + 3

x = 4/(k + 3)

That's as much as you can do with this question.

4 0

2 years ago

Tim and Jim are working on a school project together. If he had to work by himself, it would take

Vladimir [108]

Answer: 6 hours

Step-by-step explanation:

Given, Tim and Jim are working on a school project together.

Tim takes 10 hours to complete the school project.

If Jim worked alone, it would take him 15 hours

Let t be the time they both take if they work together.

Then, $\dfrac{1}{t}=\dfrac{1}{\text{Time taken by Tim}}+\dfrac{1}{\text{Time taken by jim}}$

$\Rightarrow\ \dfrac{1}{t}=\dfrac{1}{10}+\dfrac{1}{15}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{3+2}{30}=\dfrac{5}{30}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{1}{6}\\\\\Rightarrow t=6$

So, it will take<u> 6 hours</u> to complete the school project together .

4 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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2 years ago

Ming needs 1/2 pint of red paint for an art project. He has 6 jars that have the following amounts of red paint in them. He want

Dmitrij [34]

The possible answers are A C and D because they all have at least a half pint of paint.

5 0

2 years ago

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