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2 years ago

10. The expression 4 (2x-5)+7(x+2) can be written in simplest form as

Mathematics

1 answer:

ankoles [38]2 years ago

4 0

Answer:

15x-6

Step-by-step explanation: 4 * 2x = 8x

4*-5 = -20

7*x = 7x

7*2= 14

Combine like terms 8x + 7x + 14-20

15x-6

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Carrie made 127 brownies and packed 13 in each box. How many boxes are

Gala2k [10]

Answer:

9 boxes were packed. 10 brownies were left over.

Step-by-step explanation:

127 ÷ 13 = 9 R10

13 x 9 = 10 x 9 + 3 x 9 = 90 + 27 = 117

127 - 117 = 10

4 0

3 years ago

Can someone please solve this please.

Lelu [443]

Answer:

A≈576.31

Step-by-step explanation:

surface area = bh + L(s1+s2+s3)

surface area = (13*11) + 13 * (10+11+14.87)

8 0

2 years ago

Using the numbers 5, 8, and 24, create a problem using no more than 4 operations (adding, subtracting, multiplication, division,

nalin [4]

<span>√(5*8*24) or anything along those lines, you can switch the numbers around , it wont matter</span>

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3 years ago

Suppose that LMN is isosceles with base LN suppose that M

ira [324]

Given:

In an isosceles triangle LMN, LM=MN.

$m\angle M=(3x+17)^\circ,m\angle L=(2x+36)^\circ$

To find:

The measure of the angles L, M and N.

Solution:

In triangle LMN,

$LM=MN$ (Given)

$m\angle N=m\angle L=(2x+36)^\circ$ (Base angles of an isosceles triangle are equal)

Now,

$m\angle L+m\angle M+m\angle N=180^\circ$

$(2x+36)^\circ+(3x+17)^\circ+(2x+36)^\circ=180^\circ$

$(7x+89)^\circ=180^\circ$

$(7x+89)=180$

On further simplification, we get

$7x=180-89$

$7x=91$

$x=\dfrac{91}{7}$

$x=13$

The value of x is 13. Using this value, we get

$m\angle L=(2(13)+36)^\circ$

$m\angle L=(26+36)^\circ$

$m\angle L=62^\circ$

Similarly,

$m\angle M=(3(13)+17)^\circ$

$m\angle M=(39+17)^\circ$

$m\angle M=56^\circ$

And,

$m\angle N=m\angle L$

$m\angle N=62^\circ$

Therefore, the measure of angles are $m\angle L=62^\circ,m\angle M=56^\circ,m\angle N=62^\circ$ .

5 0

3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

5 0

3 years ago