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3 years ago

12

There are 104 students in fourth grade at Allison’s school.One day, 15 fourth graders were absent. How many fourth graders were

at school that day?

2 answers:

stich3 [128]3 years ago

3 0

Answer: 89 students remained at school that day.

Step-by-step explanation: If there are 104 students, and 15 are absent, you subtract, therefore, the answer is 89.

serg [7]3 years ago

3 0

89 students at school that day

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HELP PLEASE ❤️❤️!!!!!!’

strojnjashka [21]

C. The 2 in the denominator turns into -2 and you just take the coefficients directly from the numerator.

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What makes values of a and b make each inequality true? -ab>0, ab<0, a^2b>0, a/b>0

max2010maxim [7]

-ab>0 is the answer cause the others make 0 not to be simplified

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Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian

777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\B=30^o$

Applying the law of sines

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(A)}=\frac{2}{sin(30^o)}$

$sin(A)=1$

so

$A=90^o$

Find the measure of angle C

In a right triangle

we know that

$B+C=90^o$ ----> by complementary angles

$B=30^o$

therefore

$C=60^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}$

$c=2\sqrt{3}\ units$

therefore

The dimensions of the triangle are

$A=90^o$

$B=30^o$

$C=60^o$

$a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units$

<em>Second case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\A=30^o$

Applying the law of sines

Find the measure of angle B

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(30^o)}=\frac{2}{sin(B)}$

$sin(B)=0.25$

so

using a calculator

$B=14.48^o$

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$A+B+C=180^o$

$A=30^o\\B=14.48^o$

therefore

$30^o+14.48^o+C=180^o$

$C=135.52^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{a}{sin(A)}$

substitute the given values

$\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}$

$c=5.61\ units$

therefore

The dimensions of the triangle are

$A=30^o$

$B=14.48^o$

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$a=4\ units\\b=2\ units\\c=5.61\ units$

see the attached figure to better understand the problem

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Answer:

x=-8 and y=9

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distribute

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