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svlad2 [7]
3 years ago
10

Help needed please extra points given

Mathematics
1 answer:
Allisa [31]3 years ago
4 0

y^2 . w

i think you could use this

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Solve for r<br> 2r = 7/10
deff fn [24]

Answer:

7/20

Step-by-step explanation:

given,

2r = 7/10

r= 7/10*1/2

r= 7/20

6 0
3 years ago
HELP LAST ATTEMPT ( get marked brainliest show work
Sergio039 [100]

Answer:

B

Step-by-step explanation:

the side length is equal to sqrt(33)

sqrt(33)  = 5.74, closest to 5 3/4

4 0
3 years ago
Which equation is equivalent to<br><br> f(x) = 16x4 – 81 = 0?
Vesna [10]

Answer:

The equivalent expression is:

16x^4-81=(4x^2-9)(4x^2+9)

Step-by-step explanation:

We have been given the equation:

f(x)=16x^4-81=0

Equivalent expression can be computed by solving the expression ai its maximum.

We can factorize the given equation:

By using a^2-b^2=(a+b)(a-b)

Here, a=4x^2,b=9

Hence, we get the equation below:

16x^4-81=(4x^2-9)(4x^2+9)

Therefore, the equivalent expression is:

16x^4-81=(4x^2-9)(4x^2+9)

5 0
3 years ago
Read 2 more answers
Please help! The volume of an inflated spherical water balloon is 121,500 pi millimeters to the third. What is the radius of the
irga5000 [103]

Answer:

The radius of the inflated spherical balloon  is 45 millimeters.

Step-by-step explanation:

Volume of the spherical water balloon = 121,500 pi  cubic millimeters

Let the radius of the balloon = r

Now, Volume of a Sphere = \frac{4}{3} \pi r^{3}

⇒\frac{4}{3} \pi r^{3}  = 121,500 \pi

On solving for the value of r, we get:

r^{3}  = \frac{121,500\times 3}{4}   = 91125\\ \implies  r  = \sqrt[3]{91125}

or, r  = 45 millimeter

Hence, the radius of the inflated spherical balloon  is 45 millimeters.

5 0
3 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
3 years ago
Read 2 more answers
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