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3 years ago

Can someone please help me with math for college readiness

Mathematics

2 answers:

frez [133]3 years ago

8 0

It is best to start by converting all to decimals. so 7 5/6= 7.8333 and 31/4=7.75
o in order of least to greatest it is 6.79, 7.729, 31/4, 7 5/6.

bixtya [17]3 years ago

6 0

6.79,7.729,31/4,7+5/6

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I need help answering these two questions . This is missing sides and Angles of Right Triangles

iren [92.7K]

1) Tanθ= opposite/adjacent
θ=Tan-1(9/7)
θ=52.125 = 52 degrees

2) Cosθ= adjacent/hypothesis
θ=Cos-1(5/12)
θ=65.376 = 65 degrees

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2 years ago

Greatest common factor of the following monomials 40a^3b and 24a^3b^5

timofeeve [1]

the gcf is product of gcf each factor,

gcf(40a³b,24a³b⁵)

= gcf(40,24)•gcf(a³,a³)•gcf(b,b⁵)

= 8a³b

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2 years ago

Simplify the following surd expressions a) 7/3 - 2/3 + V3 - 3V3

Nata [24]

Answer:

$\frac{7}{3}-\frac{2}{3}+\sqrt{3}-3\sqrt{3}=\frac{5}{3}-2\sqrt{3}$

Step-by-step explanation:

Given the expression

$\frac{7}{3}\:-\:\frac{2}{3}\:+\:v^3\:-\:3v^3$

solving the expression

$\frac{7}{3}\:-\:\frac{2}{3}\:+\:\sqrt{3}-3\sqrt{3}$

combine the fractions i.e $\frac{7}{3}-\frac{2}{3}=\frac{5}{3}$

$\frac{7}{3}\:-\:\frac{2}{3}\:+\:\sqrt{3}-3\sqrt{3}=\frac{5}{3}+\sqrt{3}-3\sqrt{3}$

add similar elements i.e $\sqrt{3}-3\sqrt{3}=-2\sqrt{3}$

$=\frac{5}{3}-2\sqrt{3}$

Thus,

$\frac{7}{3}-\frac{2}{3}+\sqrt{3}-3\sqrt{3}=\frac{5}{3}-2\sqrt{3}$

4 0

2 years ago

Joshua finds the perimeter of the following composite figure composed of a square and right triangle so that a braid may be cut

Bumek [7]

The perimeter of a shape is the sum of its side lengths.

Joshua miscalculated the third length of the triangle

First, we calculate the third length (x) of the triangle.

Because the triangle is right-angled, we can make use of Pythagoras.

So we have:

$x^2 = 7^2 + 7^2$

$x^2 = 2(7^2)$

Take square roots of both sides

$x = \sqrt{2(7^2)$

$x = 7\sqrt{2$

The other shape in the figure is a square.

So, the perimeter (P) is

$P = 7 + x + x + x + 7$ ---- i.e. the sum of the visible lengths

So, we have:

$P = 7 + 7\sqrt2 + 7\sqrt2 + 7\sqrt2+ 7$

Evaluate like terms:

$P = 14 + 21\sqrt2$

Hence, Joshua's error is that:

He miscalculated the third length of the triangle