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3 years ago

I have a lot of algebra problems who can help i will fan medal

1 answer:

4 0

I think I can! I recently covered algebra in 6th grade, (if that's what grade ur in) and I'm pretty good at 6th grade algebra

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A label is placed around the juice box below, not including the bases. Determine the surface area of the label.

il63 [147K]

Answer:

7.5in explanation:4in+2in+1.5in=7.5in

5 0

2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 mill

ki77a [65]

Based in your question where as the oil is pumped continuously from the well at a rate proportional to the amount of the oil left in the well. so the answer of the following questions are:
A.1 is to 200,000
B.5 years

6 0

2 years ago

What is 200 +90 + 3 in standard form?

xxTIMURxx [149]

Answer: 293

200 = 2

90 = 9

3 = 3

Answer is 293.

Hope this helps!

4 0

3 years ago

2 questions in the 2 pictures and then im done thank you

Tamiku [17]

Answer:

Hey there!

For the first question we use the triangle area formula: 1/2bh, or 1/2(8)(5). This gives us 20 for the area.

For the second question we get a trapezoid, since the cross section does not pass through the vertex.

Hope this helps :)

3 0

3 years ago

PLEASE PLEASE HELP! Find the measure of APB^.

Whitepunk [10]

Answer:

65°

Step-by-step explanation:

Radii CA and CB are perpendicular to tangent lines AT and BT, so

$\angle CAT=\angle CBT=90^{\circ}$

Since angle BAT is equal to 65°, angle CAB has measure

$\angle CAB=90^{\circ}-65^{\circ}=25^{\circ}.$

Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus

$\angle CBA=\angle CAB=25^{\circ}$

The sum of the measures of all interior angles in triangle is always 180°, so

$\angle CAB+\angle CBA+\angle ACB=180^{\circ}\\ \\25^{\circ}+25^{\circ}+\angle ACB=180^{\circ}\\ \\\angle ACB=130^{\circ}$

Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so

$x=\angle APB=\dfrac{1}{2}\cdot 130^{\circ}=65^{\circ}$

8 0

3 years ago