Answer:
a = 1
Step-by-step explanation:
Right triangle DFG is congruent to right triangle DFE by the HA theorem. Corresponding parts are congruent, so ...
DG ≅ DE
a +5 = 6a
5 = 5a . . . . . . . subtract a
1 = a . . . . . . . divide by 5
Answer:
and in interval notation.
Step-by-step explanation:
We have been given a compound inequality . We are supposed to find the solution of our given inequality.
First of all, we will solve both inequalities separately, then we will combine both solution merging overlapping intervals.
Dividing by negative number, flip the inequality sign:
Dividing by negative number, flip the inequality sign:
Upon merging both intervals, we will get:
Therefore, the solution for our given inequality would be and in interval notation.
<span>In this item, we are given that out of 12 ounces of ingredients comprising the milkshake, there are 4 ounces vanilla. Expressing this into fraction will give us an answer of 4/12. This fraction can be one of the answers. Since we are asked for two fractions, we can simplify 4/12 such that we get an equivalent fraction that is equal to 2/6.
Hence, the answers to this question are 4/12 and 2/6.</span>
Answer:
The answer should be A. .
Step-by-step explanation:
Start with the second inequality, which is more straightforward than the first. Add to both sides of this equation to separate from the numbers:
. The direction of the inequality sign shall stay the same.
.
On the other hand, simplifying the first inequality takes a bit more work. Start by dividing both sides by . However, keep in mind that dividing or multiplying both sides of an inequality with a negative number will invert the direction of the inequality sign. For example, the left-hand side of an inequality might have been greater than the right-hand side. When both sides are multiplied with a number less than zero, the new left-hand side would become smaller than the new right-hand side.
Because is a negative number, dividing both sides of the first inequality inequality by will turn the "greater-than" sign into a "less-than" sign.
The inequality used to be:
.
After dividing both sides by the negative number , it becomes:
.
.
.
The question connected the two original inequalities with "or". Therefore, the conditions will be satisfied as long as either of at least one of the two inequalities is true. Note, that is true whenever is true. Therefore, any real number that is smaller than will meet the requirements- not just those that are smaller than .
The corresponding interval will be .