All categories

3 years ago

Find the sum of 16y² and -5y²

Mathematics

1 answer:

nalin [4]3 years ago

4 0

Answer:

11y²

Step-by-step explanation:

16y² + -5y²

So in this expression ignore y² and solve 16 + -5

Since both terms are like terms you can add

16 + -5

= 11

____________________________________________________________

Now add y² to your answer

= 11y²

You might be interested in

in the given figure ,o is the centr of the circle, ab is the diameter and do is perpendicular to ab. prove that angle AEC= ANGLE

maksim [4K]

The given triangles ΔDOA and ΔABC are right triangles that have a

common vertex at point A.

ΔDOA is similar to ΔABC, and ∠AEC is equal to ∠ABC, therefore, ∠AEC = ∠ODA

Reasons:

The given parameters are;

The diameter of the circle with center O = AB

DO ⊥ AB (DO is perpendicular to AB)

Required:

Prove that ∠AEC = ∠ODA

A two column proof is presented as follows;

Statement ${}$ Reason

1. AB is the diameter of circle ${}$ 1. Given

2. DO is perpendicular to AB ${}$ 2. Given

3. ∠DOA = 90° ${}$ 3. Definition of DO ⊥ AB

4. ∠BCA = 90° ${}$ 4. Thales theorem

5. ∠BCA ≅ ∠BCA ${}$ 5. Reflexive property

6. ΔDOA ~ ΔABC ${}$ 6. AA similarity postulate

7. ∠ABC ≅ ∠ODA ${}$ 7. CASTC

8. ∠ABC = ∠ODA ${}$ 8. Definition of congruency

9. ∠AEC ≅ ∠ABC ${}$ 9. Angles in the same segment

10. ∠AEC = ∠ABC   ${}$ 10. Definition of congruency

11. ∠AEC = ∠ODA ${}$ 11. Transitive property of equality

In statement 6, ΔDOA is similar to ΔABC by Angle-Angle, AA, similarity

postulate, therefore, the three angles of ΔDOA are congruent to the three

angles of ΔABC.

Therefore ∠ABC ≅ ∠ODA by Corresponding Angles of Similar Triangles

are Congruent, CASTC.

Learn more about circle theorem here:

brainly.com/question/16879446

7 0

2 years ago

What is the y intercept of (-4,6) and (1,-9)

vladimir1956 [14]

Answer:

what is the y intercept of (-4,6) and (1,-9) : (3,-12)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

40.00 jeans 20% discount

Svetllana [295]

20(40)
$\frac{20}{100}$ = 0.2
0.2(40) = 8

After price: $32.00
But, you save: $8

4 0

2 years ago

The results of joes survey show that baseball is the favorite sport for 18 out of 25 people. What percent do not like sports?

charle [14.2K]

well if 18 out of 25 like it, then the remaining 7 doesn't like it, namely 7 out of 25.

if we take 25 to be the 100%, what is 7 off of if in percentage?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 25&100\\ 7&x \end{array}\implies \cfrac{25}{7}=\cfrac{100}{x}\implies 25x=700\implies x=\cfrac{700}{25}\implies x=28$

7 0

2 years ago

When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode

yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are central tendency measures in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a normal distribution, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a distribution is not symmetrical, we say it is skewed. The skewness is a measure of the asymmetry of the distribution. In this case, the mean, median and mode are not the same, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (negative skew), or greater than them (positive skew), or approximately equal than the median but much greater than the mode (a variation of a positive skew case).

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a mound-shaped symmetrical distribution, it resembles the normal distribution and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, they are all approximately equal. So, the general relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, considering they have similar characteristics of the normal distribution, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0

2 years ago