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MArishka [77]
3 years ago
7

Please help with #7! Thanks.

Mathematics
1 answer:
mixas84 [53]3 years ago
7 0
The answer is:  9.5 meters (by visual inspection of the image (graph) attached.
_______________________________________________
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I need the answer fast pls⇒
lana66690 [7]

Answer:

530

Step-by-step explanation:

870-340 = 530

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5 0
3 years ago
What is the​ slope-intercept form of the equation of the​ line?
vovikov84 [41]
Slope intercept form is y=mx+b
8 0
3 years ago
What’s the correct answer for this?
Nitella [24]

Answer:

6 pounds and .125 ounces

Step-by-step explanation:

6 0
2 years ago
The equation (x + 6)^2 + (y + 4)^2 = 36 models the position and range of the source of a radio signal. Describe the position of
Ilia_Sergeevich [38]

Answer:

position: (-6, -4)

range: 6

Step-by-step explanation:

The equation is that of a circle centered at (-6, -4) with a radius of √36 = 6. We presume that the "position" is that of the circle's center, and the "range" is the radius of the circle.

___

The standard form equation of a circle with center (h, k) and radius r is ...

(x -h)^2 +(y -k)^2 = r^2

Matching parts of the equation, we find ...

h = -6, k = -4, r = √36 = 6.

7 0
2 years ago
Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA
AleksAgata [21]

1. Start with ΔCIJ.

  • ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
  • the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
  • ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

  • ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
  • ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
  • the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
  • ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.

3 0
2 years ago
Read 2 more answers
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