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emmainna [20.7K]
3 years ago
9

Math help lol say hi point for lolll

Mathematics
1 answer:
Levart [38]3 years ago
5 0
15/20
4*5=20
3*5=15
Therefore it would be 15/20
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Evaluate.
Radda [10]

Step-by-step explanation:

-| 12/3 - 1| 2-(-3)

-|12-3/3|5

-5|9/3|

-5|3|

-5×3 & -5×-3

-15 & 15

I think it is the answer

6 0
1 year ago
Answer 5, 6 , and 7 by today please.
kherson [118]

Answer:

Step-by-step explanation:

5. a) ∠1 and ∠2 are remote interior angles of ∠ACD so that means that ∠ACD = ∠1 + ∠2

   b) Because an exterior angle is the sum of its two remote interior angles it makes sense that an exterior angle is greater in measure than either of its remote interior angles.

6. BD = DB  Reflexive property

    ∠3 = ∠5, ∠4 = ∠6  Alt. int. angles

    ΔADB = ΔCDB   ASA

7. AB = BC Def. of midpoint

   ∠1 = ∠2 Given

   ∠BAE = ∠CBD Corresponding angles

   ΔBAE = ΔCBD ASA

    ∠D = ∠E CPCTC

3 0
2 years ago
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

5 0
3 years ago
Help again please thank you!
-Dominant- [34]
2 one I think f//said
6 0
2 years ago
A transformation T : (x, y) → (x + 3, y + 1). Find the preimage of the point (4, 3) under the given transformation. (7, 4) (1, 2
motikmotik

Answer:

(1, 2)

Step-by-step explanation:

Remember that the final shape and position of a figure after a transformation is called the image, and the original shape and position of the figure is the pre-image.

In our case, our figure is just a point. We know that after the transformation T : (x, y) → (x + 3, y + 1), our image has coordinates (4, 3).

The transformation rule T : (x, y) → (x + 3, y + 1) means that we add 3 to the x-coordinate and add 1 to the y-coordinate of our pre-image. Now to find the pre-image of our point, we just need to reverse those operations; in other words, we will subtract 3 from the x-coordinate and subtract 1 from the y-coordinate.

So, our rule to find the pre-image of the point (4, 3) is:

T : (x, y) → (x - 3, y - 1)

We know that the x-coordinate of our image is 4 and its y-coordinate is 3.

Replacing values:

                (4 - 3, 3 - 1)

                (1, 2)

We can conclude that our pre-image is the point (1, 2).

6 0
2 years ago
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