All categories

4 years ago

-3c=15 c = __ help plz

1 answer:

7 0

You would divide negative 3 (-3) from both sides to keep the c by itself. When you divide a positive by a negative it becomes negative.

Dividing 15 by -3 would equal -5, so c = -5

You might be interested in

Consider the two triangles. How can the triangles be proven similar by the SSS similarity theorem? Show that the ratios are equi

shusha [124]

Side-angle-side similarity means that when two triangles have corresponding angles that are congruent and corresponding sides with identical ratios we can say that the triangles are similar. So UV/XY and WV/ZY are equivalent and therefore, angle V equals angle Y is the correct answer.

3 0

3 years ago

Read 2 more answers

Select the correct answer.

Anit [1.1K]

Answer:

C. 21°

Step-by-step explanation:

In triangle ABC, perpendiculars dropped on the sides AB and AC from the ray BD are equal in measure or congruent.

Therefore, ray BD is equidistant from the sides AB and AC.

So, ray BD will be the bisector of angle ABC.

$\therefore m\angle ABD = m\angle DBC$

(Angle bisector theorem)

$\therefore (3x+9)\degree = (45-6x)\degree$

$\therefore 3x+9 = 45-6x$

$\therefore 3x+6x = 45-9$

$\therefore 9x = 36$

$\therefore x =\frac{36}{9}$

$\therefore x =4$

$\therefore (3x+9)\degree = (3*4+9)\degree=(12+9)\degree =21\degree$

$\implies m\angle ABD=21\degree$

4 0

3 years ago

What is the answer ?

k0ka [10]

Answer:

Step-by-step explanation:

5 0

3 years ago

Find the perimeter of the rectangle

kvv77 [185]

Answer:

Step-by-step explanation:

5+5=10

2+2=4

and all together that equals 14

8 0

3 years ago

Rationalize the denominator: GRADE 9 CBSE

kakasveta [241]

Answer:

1. a)

$\begin{gathered} \frac{2}{ \sqrt{3} - 1 } = \frac{2}{ \sqrt{3} - 1} \times \ \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } \\ = \frac{2 \sqrt{3} + 2 } {( \sqrt{3} ) {}^{2} - 1 {}^{2} } = \frac{2 \sqrt{} 3}{3 - 1} = \frac{2 \sqrt{3} }{2} \\ = \sqrt{3} \end{gathered}$

$\\$

$\begin{gathered} = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \\ \\ = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \times \frac{ \sqrt{12} + \sqrt{5} }{ \sqrt{12} + \sqrt{5} } \\ \\ = > \: \: \frac{7( \sqrt{12} + \sqrt{5} ) }{ {( \sqrt{12}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ = > \: \: \frac{7( \sqrt{12} + \sqrt{5}) }{12 - 5} \\ \\ => \: \: \frac{ \cancel{7}( \sqrt{12} + \sqrt{5} ) }{ \cancel{7}} \\ \\ => \: \: \sqrt{12} + \sqrt{5} \end{gathered}$

7 0

2 years ago