What two simler problems can you use to find 9 x 38?(Hint: think about ones and tens.)

2 answers:

6 0

Nine times thirty-eight can also be thirty-eight times nine or (18 divided by 2) multiplied by 38 or like this

(18/2)times(38/2)

I hope this gives you an idea

julia-pushkina [17]3 years ago

3 0

10 x 38 - 38= Multiplied by 10 just add a 0 380 Now subtract the extra 38 342

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reflect (7,4) across the x-axis then reflect the result across the y-axis what are the coordinates of the final point

Andrews [41]

The coordinates of the final point after the two reflections is (-7, -4)

<h3>How to reflect a point across an axis?</h3>

For a point (x, y), a reflection across the x-axis gives the point (x, -y), while for a point (x, y) a reflection across the y-axis gives (-x, y).

In this case, we have the point (7, 4), first we reflect across the x-axis, so we get the point (-7, 4).

Now we reflect across the y-axis, so the sign of the y-component changes, and we get (-7, -4)

So the coordinates of the final point after the two reflections is (-7, -4).

If you want to learn more about reflections, you can read:

brainly.com/question/4289712

8 0

2 years ago

Find the 20th term in the sequence, where the first term is -2 and the common ratio is-2.

geniusboy [140]

Answer:

Step-by-step explanation:

(1)

a1=-2

r=-2

$a_{20}=(-2)(-2)^{20-1}=-2(-2)^{19}=2^{20}=1048576\\(2)\\a_{1}=1000\\r=\frac{200}{1000} =\frac{1}{5} \\a_{8}=1000(\frac{1}{5} )^{8-1}=1000(.2)^7=0.0128\\(3)\\a_{1}=2\\a_{3}=50\\a_{2}=\sqrt{2*50} =10\\r=\frac{10}{2} =5$