Y=6x (1)
y=5x-7 (2)
Substitute y into (2)
(6x)=5x-7 -- subtract 5x from both sides
x=-7
Sub x into 1
y=6(-7)
y=-42
x=-7
y=-42
Answer:
The correct option is B.
Step-by-step explanation:
Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .
Draw a diagonal AC.
In triangle BCA and DAC,
AC\cong ACAC≅AC (Reflexive Property of Equality)
\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)
\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)
The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.
By ASA postulate,
\triangle BCA\cong \triangle DAC△BCA≅△DAC
Therefore option B is correct
<span>inversely
y = k/x
k = yx
k = 10*10 = 100
when y = 20
20 = 100/x
20x = 100
x = 5
answer
</span><span> x = 5 when y is 20</span>
For this case, the first thing we are going to do is assume that all the tests are worth the same.
Then, we define a variable:
x: score of Mona's last test
We write now the inequality that models the problem:

From here, we clear the value of x:
Answer:
the lowest grade that Mona can get for her last test so that her test average is 90 or more is:
x = 87