The complete version of question:
<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>
Answer:
![5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}](https://tex.z-dn.net/?f=5%5Cleft%28x%2B27%5Cright%29%5Cge%20%5C%3A6%5Cleft%28x%2B26%5Cright%29%5Cquad%20%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ax%5Cle%20%5C%3A-21%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%2C%5C%3A-21%5D%5Cend%7Bbmatrix%7D)
Step-by-step explanation:
As the description of the statement is:
'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>
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As
- <em>Five times the sum of a number and 27 </em>is written as:
- <em>greater than or equal </em>is written as:
- <em>six times the sum of that number and 26' </em>is written as: 6(x + 26)
so lets combine the whole statement:
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solving
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Therefore,
9514 1404 393
Answer:
∠CAB = 28°
∠DAC = 64°
Step-by-step explanation:
What you do in each case is make use of the relationships you know about angles in a triangle and around parallel lines. You can also use the relationships you know about diagonals in a rectangle, and the triangles they create.
<u>Left</u>
Take advantage of the fact that ∆AEB is isosceles, so the angles at A and B in that triangle are the same. If we call that angle measure x, then we have the sum of angles in that triangle is ...
x + x + ∠AEB = 180°
2x = 180° -124° = 56°
x = 28°
The measure of angle CAB is 28°.
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<u>Right</u>
Sides AD and BC are parallel, so diagonal AC can be considered a transversal. The two angles we're concerned with are alternate interior angles, so are congruent.
∠BCA = ∠DAC = 64°
The measure of angle DAC is 64°.
(Another way to look at this is that triangles BCE and DAE are congruent isosceles triangles, so corresponding angles are congruent.)
In order to solve this, we need to figure out how much ONE student spends.
dollars per student.
If 10 student spent for lunch, we just have to multiply our number by how much each student spends.
