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VladimirAG [237]
3 years ago
5

What is the measure of ABD?

Mathematics
1 answer:
Lostsunrise [7]3 years ago
3 0

Answer:

try d or c

Step-by-step explanation:

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postnew [5]

\\ \sf\longmapsto \dfrac{(0.0225)^{\frac{-3}{2}}\times (0.0001)^{\frac{3}{4}}}{(0.0125)^{\frac{1}{3}}}

\\ \sf\longmapsto \dfrac{((0.15)^2)^{\frac{-3}{2}}\times ((0.1)^4)^{\frac{3}{4}}}{(0.5)^3)^{\frac{1}{3}}}

\\ \sf\longmapsto \dfrac{(0.15)^{2\times \dfrac{-3}{2}}\times (0.1)^{4\times \dfrac{3}{4}}}{(0.5)^{3\times \dfrac{1}{3}}}

\\ \sf\longmapsto \dfrac{(0.15)^{-3}(0.1)^3}{(0.5)^1}

\\ \sf\longmapsto \dfrac{\dfrac{1}{(0.15)^3}\times 0.001}{0.5}

\\ \sf\longmapsto \dfrac{\dfrac{1}{0.003375}(0.001)}{0.5}

\\ \sf\longmapsto\dfrac{ \dfrac{1000000}{3375}\times \dfrac{1}{1000}}{0.5}

\\ \sf\longmapsto \dfrac{\dfrac{1000}{3375}}{0.5}

\\ \sf\longmapsto \dfrac{1000}{3375}\times \dfrac{10}{5}

\\ \sf\longmapsto \dfrac{2000}{3375}

\\ \sf\longmapsto 0.592

4 0
2 years ago
-10 + 4<br><br><br>plz heeeeeelp
zimovet [89]

Answer:

-6

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Which equation represents a circle whose center is located at (- 6, 2) and whose radius is sqrt(10) units?
azamat

Answer:

{(x  + 6)}^{2}  +  {(y - 2)}^{2}=10

Step-by-step explanation:

The equation of a circle with center (a,b) and radius r, is

{(x - a)}^{2}  +  {(y - b)}^{2}  =  {r}^{2}

From the question, the center of the circle is located at (-6,2) and the radius is

r =  \sqrt{10}

We substitute the center and radius to get:

{(x -  - 6)}^{2}  +  {(y - 2)}^{2}  =  {( \sqrt{10} })^{2}

Simplify to obtain:

{(x  + 6)}^{2}  +  {(y - 2)}^{2}=10

7 0
2 years ago
BC and DF are parallel lines. B is a point on AD and E is a point of DF. BD=BE. work out angle DBE give a reason. ITS ON MATHSWA
zloy xaker [14]

Answer:

(Angle DBE) ∠DBE = 50°

Step-by-step explanation:

It is crucial for us to be able to interpret plane geometry related question in order to solve them efficiently.

From the question given; we have analysed and come up with a diagrammatic expression and mathematical solution that clearly explains the question .

Given that

BC and DF are parallel lines.

B is a point on AD  

B is a point on AD

BD=BE

work out angle DBE and give a reason.

The diagram can be seen in the attached image below.

SInce ∠BD = ∠DE

Then triangle BDE is an isosceles triangle. In an isosceles triangle ; two sides are equal in length.

The sum of angles in an isosceles triangle = 180° (sum of angles in a triangle)

So;

∠BDE + ∠BED + ∠DBE = 180° (sum of angles in a triangle)

From the diagram; we will see that ∠ABC = ∠BDE  (corresponding angle)

Since ; ∠ABC is not given and which is needed to solve this question.

Let's just assume that ∠ABC is 65° , the main thing is to be able to interpret and understand the concept of the question.

Now;

Since

∠ABC = ∠BDE  

∠ABC  = 65°

∠BDE  = 65°

Again;

∠BDE + ∠BED + ∠DBE = 180° (sum of angles in a triangle)

∠BDE will be aso equal to  ∠BED ; this is because since the length of the opposite sides of the isosceles triangle are equal, their angles will also be equal.

Therefore;

65° + 65° + ∠DBE = 180° (sum of angles in a triangle)

130 ° + ∠DBE = 180° (sum of angles in a triangle)

∠DBE = 180° - 130°

∠DBE = 50°

3 0
3 years ago
Sandra throws an object straight up into the air with initial velocity of 38 ft per square from platform that is 30 ft above the
expeople1 [14]

Answer:

3 seconds.

Step-by-step explanation:

We have been given that Sandra throws an object straight up into the air with initial velocity of 38 ft per second from platform that is 30 ft above the ground.

To find the time it will take for object to hit the ground, we will use formula:  

h(t)=-16t^2+v_0t+h_0, where, v_0 represents initial velocity and h_0 represents initial height.

Upon substituting our given initial velocity and initial height, we will get:

h(t)=-16t^2+38t+30

We know that object will hit the ground, when height will be 0, so we will equate object's height equal to 0 and solve for t as:

-16t^2+38t+30=0

Divide by 2:

-8t^2+19t+15=0

Upon using quadratic formula, we will get:

t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-19\pm\sqrt{19^2-4(-8)(15)}}{2(-8)}

t=\frac{-19\pm\sqrt{361+480}}{-16}=\frac{-19\pm\sqrt{841}}{-16}

t=\frac{-19\pm29}{-16}

t=\frac{-19-29}{-16},t=\frac{-19+29}{-16}

t=\frac{-48}{-16},t=\frac{10}{-16}

t=3,t=-\frac{5}{8}

Since time cannot be negative, therefore, it will take 3 seconds for object to hit the ground.

7 0
2 years ago
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