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3 years ago

Give the digits in the ones place and the hundredths place. 53.19

1 answer:

4 0

The ones place is 3 and the hundredths place is 9. Always remember, on the left side of the point starts with ones followed be tens, but in the right side of the point it starts with tenths followed by hundredth.

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Find 2 numbers who product is 15 and sum is 16.

Marizza181 [45]

Answer:

15 and 1

Step-by-step explanation:

x and y are two numbers.

Two equations:

x · y = 15

x + y = 16

Rearrange one of the equations (I'll rearrange the sum equation):

x + y = 16

x = 16 - y

Substitute that to the other equation and solve for y:

x · y = 15

(16 - y) · y = 15

16 - y · y = 15

16 - y² = 15

-y² = 15 - 16

-y² = -1

y² = 1

y = √1

y = 1

Now substitute that to any of the equation and solve for x (in here, I'll choose the multiplication one):

x · y = 15

x · 1 = 15

x = 15

Now verify:

15 · 1 = 15

15 + 1 = 16

This is correct

6 0

3 years ago

If you use squares to approximate area,what do you use to approximate volume?

erik [133]

Answer:

cubes

Step-by-step explanation:

because when you find area you would say 2 meters "squared" but when you find volume you woukd say 2 meters "cubed"

3 0

3 years ago

NEED HELP!!! WILL MARK BRAINLIEST!!!

Tanzania [10]

Answer:

RTA= (x-2)·(x-11/2)/(x-2)(x-4)= (you can simplify again if you want by eliminating both (x-2)

(x-11/2)/(x-4)

Step-by-step explanation:

Ok we need to simplify the expression so:

x^2+3x-10= Bhaskara formula=

-3(±√9-4·1·(-10))/2·1=

X1=(-3+7/2)--> X1=2(R)---> (X-R)--->X-2

X2=(-3-7)/2 --> X2=11/2(R)---> (X-R)--->X-11/2

x^2-6x+8= Bhaskara formula=

6(±√36-4·1·8)/2·1=

X1=(6-2)/2=2--> X1=2(R)---> (X-R)--->X-2

X2=(6+2)/2=2--> X2=4(R)---> (X-R)--->X-4

so, The simplify expression is

(x-2)·(x-11/2)/(x-2)(x-4)=

8 0

3 years ago

Please help!!

Elodia [21]

Answer:

$AC=\sqrt{13}$

Step-by-step explanation:

As the the length of the segments BC = AC, so the triangle ABC must be an isosceles triangle.

As CD ⊥ AB, it means the base AB gets bisected by CD, making two congruent angles named as ∠ACD and ∠BCD. As shown in attached figure.

CD bisecting the base AB into two equal parts means if AB = 4, then

the length of AD = 2 and the length of DB = 2. Hence, AD = DB

As BC = AC, it means these two equal sides of isosceles triangle would make equal angles on the opposite side. Hence, angle ∠A and ∠B would be equal as shown in attached figure.

As CD is bisecting the base AB into two equal parts. Hence, it can be treated as the part of two right triangles named ACD and BCD.

So, the length of AC can be easily determined using using the Pythagoras formula.

So,

c² = a² + b²

$c=\sqrt{a^{2}+b^{2}}$