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2 years ago

Can somebody help me solve this?

Mathematics

1 answer:

Nookie1986 [14]2 years ago

3 0

Answer:

z = 3.2 units

Step-by-step explanation:

sin 76°/z = sin 51°/2.6

cross-multiply to get:

z·(sin 51°) = 2.6·(sin 76°)

z = 2.6·(sin 76°) ÷ (sin 51°)

z = 3.2

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The shaded region in the graph represents the solutions of this system of linear inequalities:

kotegsom [21]

Answer:

$y$

$y\leq 2x+3$

Step-by-step explanation:

step 1

Find the equation of the solid line

From the graph take the points (0,3) and (4,11)

Find the slope

$m=(11-3)/(4-0)\\m=8/4=2$

The equation of the solid line in slope intercept form is equal to

$y=mx+b$

we have

$m=2$

$b=3$ ----> the y-intercept is the point (0,3)

substitute

$y=2x+3$

therefore

The inequality is

$y\leq 2x+3$

step 2

Find the equation of the dashed line

The slope is given

$m=4$

From the graph take the y-intercept (0,-5)

The equation of the solid line in slope intercept form is equal to

$y=mx+b$

we have

$m=4$

$b=-5$

substitute

$y=4x-5$

therefore

The inequality is

$y$

because the shaded region is below the dashed line

therefore

The system of inequalities is

$y$

$y\leq 2x+3$

8 0

2 years ago

*25 points*

maxonik [38]

Answer:

Step-by-step explanation:

root the 216 as it is a cube

:edge *edge *edge =volume

Since all edges are the same, just root it

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1 year ago

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Step-by-step explanation:

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2 years ago

Can 15x-10 be factored

Anna [14]

Yes because removing the common factor is important. There are a few steps, then you can factor by regrouping. Hope this helps.

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3 years ago

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Use a paragraph, flow chart, or two-column proof to prove the angle congruency. Given: ∠ CXY ≅ ∠ BXY

Llana [10]

Answer:

See explanation

Step-by-step explanation:

Consider triangles CAX and BAX. In these triangles,

$\overline{AC}\cong \overline{AB}$ - given
$\angle CAX\cong \angle BAX$ - given
$\overline{AX}\cong \overline{AX}$ - reflexive property

By SAS postulate, $\triangle CAX\cong \triangle BAX$

Congruent triangles have conruent corresponding parts. So,

$\overline{CX}\cong \overline{BX}$

Consider triangles CXY and BXY. In these triangles,

$\overline{CX}\cong \overline{BX}$ - proven
$\angle CXY\cong \angle BXY$ - given
$\overline{XY}\cong \overline{XY}$ - reflexive property

By SAS postulate, $\triangle CXY\cong \triangle BXY$

Congruent triangles have conruent corresponding parts. So,

$\angle XCY\cong \angle XBY$

5 0

2 years ago