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Tom [10]
2 years ago
15

If Line segment C B. bisects ∠ACD, what additional information could be used to prove ΔABC ≅ ΔDBC using SAS? Select three option

s.
m∠ABC = 125° and AB ≅ DB
ΔACD is isosceles with base AD
ΔABD is isosceles with base AD
CD = 52 cm
AB = 29 cm

Mathematics
2 answers:
dolphi86 [110]2 years ago
7 0

Answer:

<h2>A B D</h2>

Step-by-step explanation:

meriva2 years ago
3 0

Answer:

Option (1)

Step-by-step explanation:

In the figure attached,

BC is the angle bisector of angle ACD.

To prove ΔABC and ΔDBC congruent by SAS property we require two sides and the angle between these sides to be congruent.

Since BC ≅ BC [Reflexive property]

∠ABC ≅ ∠CBD ≅ 125°

And sides AB ≅ BD

Both the triangles will be congruent.

Therefore, additional information required to prove ΔABC ≅ ΔDBC have been given in option (1).

Therefore, Option (1) will be the answer.

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How many liters of wine can be held in a wine barrel whose capacity is 28.0 gal? 1 gal = 4 qt = 3.7854 l?
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Answer:

105.9912 liters of wine can be held in a wine barrel whose capacity is 28.0 gal.

Step-by-step explanation:

Proportion states that the two ratios or fractions are equal.

Let x be the liters of wine can be held in a wine barrel whose capacity is 28.0 gal.

Given that:

1 gal = 3.7854 l

then

by definition of proportions;

\frac{1}{3.7854} =\frac{28}{x}

By cross multiply we get;

x = 28 \times 3.7854 = 105.9912 liters

Therefore, 105.9912 liters of wine can be held in a wine barrel whose capacity is 28.0 gal.

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Learning Task 3. Find the equation of the line. Do it in your notebook.
Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is y = 5\cdot x +9. The equation of the line in standard form is -5\cdot x + y = 9.

2) The equation of the line in slope-intercept form is y = \frac{2}{5}\cdot x +\frac{14}{5}. The equation of the line in standard form is -2\cdot x +5\cdot y = 14.

3) The equation of the line in slope-intercept form is y = 3\cdot x +4. The equation of the line in standard form is -3\cdot x +y = 4.

4) The equation of the line in slope-intercept form is y = 2\cdot x + 6. The equation of the line in standard form is -2\cdot x +y = 6.

5) The equation of the line in slope-intercept form is y = \frac{5}{6}\cdot x -\frac{7}{6}. The equation of the line in standard from is -5\cdot x + 6\cdot y = -7.

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: (m = 5, x = -1, y = 4)

4 = (5)\cdot (-1)+b

4 = -5 +b

b = 9

The equation of the line in slope-intercept form is y = 5\cdot x +9.

Then, we obtain the standard form by algebraic handling:

-5\cdot x + y = 9

The equation of the line in standard form is -5\cdot x + y = 9.

2) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = 3, y_{1} = 4, x_{2} = -2, y_{2} = 2)

3\cdot m + b = 4 (Eq. 1)

-2\cdot m + b = 2 (Eq. 2)

From (Eq. 1), we find that:

b = 4-3\cdot m

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

-2\cdot m +4-3\cdot m = 2

-5\cdot m = -2

m = \frac{2}{5}

And from (Eq. 1) we find that the y-Intercept is:

b=4-3\cdot \left(\frac{2}{5} \right)

b = 4-\frac{6}{5}

b = \frac{14}{5}

The equation of the line in slope-intercept form is y = \frac{2}{5}\cdot x +\frac{14}{5}.

Then, we obtain the standard form by algebraic handling:

-\frac{2}{5}\cdot x +y = \frac{14}{5}

-2\cdot x +5\cdot y = 14

The equation of the line in standard form is -2\cdot x +5\cdot y = 14.

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: (m = 3, b = 4)

y = 3\cdot x +4

The equation of the line in slope-intercept form is y = 3\cdot x +4.

Then, we obtain the standard form by algebraic handling:

-3\cdot x +y = 4

The equation of the line in standard form is -3\cdot x +y = 4.

4) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = -3, y_{1} = 0, x_{2} = 0, y_{2} = 6)

-3\cdot m + b = 0 (Eq. 3)

b = 6 (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

-3\cdot m+6 = 0

3\cdot m = 6

m = 2

The equation of the line in slope-intercept form is y = 2\cdot x + 6.

Then, we obtain the standard form by algebraic handling:

-2\cdot x +y = 6

The equation of the line in standard form is -2\cdot x +y = 6.

5) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = -1, y_{1} = -2, x_{2} = 5, y_{2} = 3)

-m+b = -2 (Eq. 5)

5\cdot m +b = 3 (Eq. 6)

From (Eq. 5), we find that:

b = -2+m

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

5\cdot m -2+m = 3

6\cdot m = 5

m = \frac{5}{6}

And from (Eq. 5) we find that the y-Intercept is:

b = -2+\frac{5}{6}

b = -\frac{7}{6}

The equation of the line in slope-intercept form is y = \frac{5}{6}\cdot x -\frac{7}{6}.

Then, we obtain the standard form by algebraic handling:

-\frac{5}{6}\cdot x +y =-\frac{7}{6}

-5\cdot x + 6\cdot y = -7

The equation of the line in standard from is -5\cdot x + 6\cdot y = -7.

6 0
1 year ago
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