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3 years ago

in rhombus ABCD, diagonals AC and BD meet at point E. If the measure of angle DAB is 46 find the length of EB

Mathematics

1 answer:

Nuetrik [128]3 years ago

3 0

Obtuse witch is 1.50 degrees and you can barley measure the veryex

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What whole number is equivalent to 8/2

svetoff [14.1K]

4 is the whole number equivalent. When dealing with an improper fraction or one that is "top heavy" you take the denominater (the bottom number) unto the numerator.
Two(2)goes into 8, 4 times, making it whole!

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3 years ago

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Please help me with 25-29!!

vredina [299]

25-29 ok so first do 9-5 equals 4

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2 years ago

4. Find the value of the expression below for r=4 and t=2.

Katen [24]

Answer:

(a) 9

Step-by-step explanation:

$\sf t^3 - r + 20 \div r$

substitute r = 4, t = 2

$\sf (2)^3 - 4 + 20 \div 4$

simplify

$\sf (2)^3 - 4 + 5$

cubic of 2 is 8

$\sf 8 - 4 + 5$

simplify

$\sf 9$

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1 year ago

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Angles ∝ and β are the two acute angles in a right triangle. Use the relationship between sine and cosine to find the value of β

arlik [135]

Ok so cos(x) = sin(90-x) so using that you can get sin(7x-15) = sin(90-(3x+5)) so 7x-15 = -3x+85. 10x = 100 and x = 10. Use that to find the 2 angles are 55 and 35

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3 years ago

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

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