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Oduvanchick [21]
2 years ago
11

I need help on saxon math algebra 1 lessons 36-40 can you please help me ,

Mathematics
2 answers:
saul85 [17]2 years ago
7 0
36-40= -4
36 subtracted by 40 equals negative 4
-4
Oksi-84 [34.3K]2 years ago
5 0

Answer:

ye asure im actually in lesson 85 right now

Step-by-step explanation:

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PLEASE HELP ASAP!! i will mark you brainliest
V125BC [204]

Answer:

1.  I am not sure about the first one. I tried everything and nothing I got from any of my work is looking right.

2. AB = 5

3. 8

4 0
3 years ago
Please, I really need help-
posledela
<h3>Answer: 48</h3>

=====================================================

Explanation:

Triangle ABC is isosceles because AC = BC.

The angles opposite these congruent sides are angle ABC and angle BAC. These are the base angles.

For any isosceles triangle, the base angles are congruent.

Angle BAC = 69 degrees is given. So angle ABC = 69 as well.

-------------------

The missing angle must add to the two other angles so that all three angles in a triangle add to 180

(angle ABC) + (angle BAC) + (angle BCA) = 180

69 + 69 + (angle BCA) = 180

138 + (angle BCA) = 180

angle BCA = 180-138

angle BCA = 42

-------------------

Angle DCE is congruent to angle BCA because they are vertical angles.

Triangle DCE is a right triangle. The missing angle is 90-(angle BCA) = 90-42 = 48

angle EDC = 48 degrees

6 0
3 years ago
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Whats the equation of the straight line on this graph<br><br> a) y=3x+4<br> b) y=2x+4<br> c) y=4x+4
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B. y=2x+4
hope this helps, have a great day!
3 0
3 years ago
This is a Differential Equations problem, I only need help on part 1 from question five. I need steps as well, thank you.
jolli1 [7]

Answer:

a) y(t) =  \sqrt{2t + 1 }

b) 1.5 hours after thickness will be 2 inches.

Step-by-step explanation:

\frac{dy}{dt}  =  \frac{1}{y}   \\   \\ ydy = dt  \\ \\ integrating \: both \: sides \\   \\  \int y \: dy =  \int 1 \: dt \\  \\  \frac{ {y}^{2} }{2}  = t + c \\  \\  {y}^{2}  = 2t + 2c \\  \\ y (t)=  \sqrt{2t + 2c} ...(1) \\  \\ a)  \:  \: \: plug \: t = 0 \: in \: (1) \\  \\  y (0)=  \sqrt{2(0) +2 c} \\  \\ y (0)=  \sqrt{0 +2 c} \\  \\ 1 =  \sqrt{2c}  \:  \: ( \because \: y (0)=1) \\  \\  2c = 1   \:  \implies \: c =  \frac{1}{2} \\  \\ plug \: c = \frac{1}{2} \: in \: (1) \\  \\ y(t) =  \sqrt{2t + 2 \times  \frac{1}{2} } \\  \\  \huge \:  \red {y(t) =  \sqrt{2t + 1 } } \\  \\b) \:  \:  plug \: y(t) = 2 \: in \: above \: equation \\  \\ 2 =  \sqrt{2t + 1}  \\  \\ 4 = 2t + 1 \: \\  (squaring \: both \: sides) \\  \\ 4 - 1 = 2t \\  \\ 2t = 3 \\  \\ t =  \frac{3}{2}  \\  \\ t = 1.5 \: hours \\

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2 years ago
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