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devlian [24]
3 years ago
9

Mike has 3 planting boxes for her flowers . each boxes is 4 feet wide and 8a feet long how much area for planting flowers does m

ike have altogether ?
Mathematics
1 answer:
ella [17]3 years ago
4 0
66 I am pretty sure hope this helps :)
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Please make original sentences with the words<br><br> -wallop<br> -diametrically <br> -luminous
Bess [88]

Answer:

-"My step-dad's punch packs a wallop!"

-"The stars shining in the night sky were luminous."

-"The two diametrically opposed viewpoints of ketchup and mustard have been on war for decades."

4 0
3 years ago
Read 2 more answers
Given: m∠ADB = m∠CDB<br><br> AD ≅ DC <br> Prove: m∠BAC = m∠BCA <br> BD⊥ AC
puteri [66]

Answer:

The proof is explained below.

Step-by-step explanation:

Given m∠ADB = m∠CDB and AD ≅ DC

we have to prove that m∠BAC = m∠BCA  and BD⊥ AC

In ΔADO and ΔCDO

∠OAD=∠OCD          (∵ADC is an isosceles triangle)

AD=DC                     (∵Given)

∠ADO=∠CDO          (∵Given)

By ASA rule, ΔADO≅ΔCDO

In ΔBAD and ΔBCD

AD=DC                (∵ABC is an isosceles triangle)

∠ADB=∠CDB      (∵Given)

DB=DB                (∵common)

By ASA rule, ΔADB≅ΔCDB

Now, ΔADB≅ΔCDB and  ΔADO≅ΔCDO

⇒  ΔADB-ΔADO≅ΔCDB-ΔCDO

⇒  ΔABO≅ΔCBO

Hence, by CPCT,  m∠BAC = m∠BCA

Now, we have to prove that BD⊥ AC i.e we have to prove m∠BOA=90°

Now, ΔABO≅ΔCBO therefore by CPCT,  m∠BOA = m∠BOC

But, m∠BOA + m∠BOC=180°   (linear pair)

⇒  m∠BOA + m∠BOA=180°

⇒  2m∠BOA=180°  ⇒  m∠BOA=90°

Hence,  BD⊥ AC

3 0
4 years ago
A right triangle is ____ a scalene triangle.<br> Always, sometimes, or never
konstantin123 [22]

Answer: Sometimes

Step-by-step explanation:

This is a very tricky question. We know that in an isosceles triangle, two sides are the same. Because of this, we know that the right angle is always 90 degrees, and the other sides would be 45 degrees each because all triangles add up to 180 degrees. A right triangle could also be scalene, meaning none of the sides are the same, for example a scalene triangle could have sides of 44, 46, and 90 degrees. Rest assured, a right triangle will never be equalateral. Hope this helps :) Merry Christmas

4 0
3 years ago
Which expression could be used to find the combined area of the right and left faces of the prism
erastovalidia [21]

Answer:

3'2+3'2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Can someone solve for x
Helen [10]

bearing in mind that, whenever we have an absolute value expression, is in effect a piece-wise function with a positive and a negative version of the expression, so

\bf |x^2-4x-5|=7\implies \begin{cases} +(x^2-4x-5)=7\\\\ -(x^2-4x-5)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x^2-4x-5)=7\implies x^2-4x-5=7\implies x^2-4x-12=0 \\\\\\ (x-6)(x+2)=0\implies x= \begin{cases} 6\\ -2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ -(x^2-4x-5)=7\implies x^2-4x-5=-7\implies x^2-4x+2=0 \\\\\\ (x-2)(x-2)=0\implies x = 2

6 0
3 years ago
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