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Sergeeva-Olga [200]
3 years ago
10

3.629 rounded to the nearest tenth?

Mathematics
2 answers:
photoshop1234 [79]3 years ago
7 0
Answer: 3.6

3.629≈3.6

Explanation: Use the slogan, “5 or above, give it a shove” to help you round to the nearest tenth.
Nastasia [14]3 years ago
5 0

Answer: Hello, 6

Therefore, the tenths value of 3.629 remains 6. The following table contains starting numbers close to 3.629 rounded to the nearest 10th.

...

Step-by-step explanation:

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Anastasy [175]

Answer:

48.01

Step-by-step explanation:

5(7)+4= 39

4(7)=28

39²+ 28²= 2305

√2305 = 48.01

4 0
3 years ago
Determine the value of y, if x is -2.<br> y = x² +8
KiRa [710]

Answer:

y=x²+8

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6 0
2 years ago
42:28
gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
Hi boys can you help me pls!!!!!!!
goblinko [34]

Answer: OPTION A.

Step-by-step explanation:

Given the following function:

h(x)=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{2}

You know that it represents the the height of the ball (in meters) when it is a distance "x" meters away from Rowan.

Since it is a Quadratic function its graph is parabola.

So, the maximum point of the graph modeling the height of the ball is the Vertex of the parabola.

You can find the x-coordinate of the Vertex with this formula:

x=\frac{-b}{2a}

In this case:

a=-\frac{1}{4}\\\\b=\frac{1}{2}

Then, substituting values, you get:

x=\frac{-\frac{1}{2}}{(2)(-\frac{1}{4}))}\\\\x=1

Finally, substitute the value of "x" into the function in order to get the y-coordinate of the Vertex:

h(1)=y=-\frac{1}{4}(1)^2+\frac{1}{2}(1)+\frac{1}{2}\\\\y=0.75

Therefore, you can conclude that:

<em> The maximum height of the ball is 0.75 of a meter, which occurs when it is approximately 1 meter away from Rowan.</em>

7 0
3 years ago
Which expression represents the number 5i4+2i3+8i2+−4  rewritten in a+bi  form?
aniked [119]
Remember
i²=-1 so
i³=-i
i⁴=1

so
5i⁴+2i³+8i²-4
5(1)+2(-i)+8(-1)-4
5-2i-8-4
-7-2i is the form
8 0
3 years ago
Read 2 more answers
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