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2 years ago

Hurry im running out of time

Mathematics

1 answer:

Dimas [21]2 years ago

3 0

Answer:

it is ∠EBA and ∠DBC i think

Step-by-step explanation:

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HELP HELP ME PLZ !!!

nikklg [1K]

Answer:

GK=JK

Step-by-step explanation:

SSS congruency is side-side-side congruency. We are given that GH=JH and KH=KH, which are both sides.

For SSS, we need 3 pairs of congruent sides. We already have 2 pairs.

The options are:

<G=<J

<H=<H

GK=JK

the first two options, <G=<J and <H=<H are talking about congruent angles. We don't need to know about congruent angles for SSS

GK=JK is talking about a pair of congurent sides. Then, we would have 3 pairs of congruent sides, satisfying the criteria needed for SSS. Therefore, we must know that GK=JK.

Hope this helps!

3 0

2 years ago

Multiply (-6 + i)(-2 + 9i).

malfutka [58]

Answer:

Step-by-step explanation:

go with that and you may win at this multiply question

4 0

2 years ago

Read 2 more answers

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$